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How to Calculate Combinations

Last Updated: January 16, 2024 References

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 132,720 times. Learn more...

Permutations and combinations have uses in math classes and in daily life. Thankfully, they are easy to calculate once you know how. Unlike permutations , where group order matters, in combinations, the order doesn't matter. [1] X Research source Combinations tell you how many ways there are to combine a given number of items in a group. To calculate combinations, you just need to know the number of items you're choosing from, the number of items to choose, and whether or not repetition is allowed (in the most common form of this problem, repetition is not allowed).

Calculating Combinations Without Repetition

Step 1 Consider an example problem where order does not matter and repetition is not allowed.

  • For instance, you may have 10 books, and you'd like to find the number of ways to combine 6 of those books on your shelf. In this case, you don't care about order - you just want to know which groupings of books you could display, assuming you only use any given book once.

{}_{{n}}C_{{r}}

  • If you have a calculator available, find the factorial setting and use that to calculate the number of combinations. If you're using Google Calculator, click on the x! button each time after entering the necessary digits.
  • For the example, you can calculate 10! with (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800. Find 4! with (4 * 3 * 2 * 1), which gives you 24. Find 6! with (6 * 5 * 4 * 3 * 2 * 1), which gives you 720.
  • Then multiply the two numbers that add to the total of items together. In this example, you should have 24 * 720, so 17,280 will be your denominator.
  • Divide the factorial of the total by the denominator, as described above: 3,628,800/17,280.
  • In the example case, you'd do get 210. This means that there are 210 different ways to combine the books on a shelf, without repetition and where order doesn't matter.

how to solve how many combinations problems

To calculate the number of r-combinations from a set of n elements, we use the binomial coefficient notation C(n,r), which gives the formula C(n,r) = n! / (r!(n-r)!). This formula counts the number of ways to choose an unordered subset of r elements from a set of n elements. For example, say we want to know the number of ways to pick a committee of 5 people from a group of 12. Here, n=12 and r=5. Plugging into the formula, we get C(12,5) = 12! / (5!(12-5)!) = 792.

Calculating Combinations with Repetition

  • For instance, imagine that you're going to order 5 items from a menu offering 15 items; the order of your selections doesn't matter, and you don't mind getting multiples of the same item (i.e., repetitions are allowed).

{}_{{n+r-1}}C_{{r}}

  • This is the least common and least understood type of combination or permutation, and isn't generally taught as often. [9] X Research source Where it is covered, it is often also known as a k -selection, a k -multiset, or a k -combination with repetition. [10] X Research source

{}_{{n+r-1}}C_{{r}}={\frac  {(n+r-1)!}{(n-1)!r!}}

  • If you have to solve by hand, keep in mind that for each factorial , you start with the main number given and then multiply it by the next smallest number, and so on until you get down to 0.
  • For the example problem, your solution should be 11,628. There are 11,628 different ways you could order any 5 items from a selection of 15 items on a menu, where order doesn't matter and repetition is allowed.

Community Q&A

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how to solve how many combinations problems

  • ↑ https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php
  • ↑ https://betterexplained.com/articles/easy-permutations-and-combinations/
  • ↑ https://www.mathsisfun.com/combinatorics/combinations-permutations.html
  • ↑ https://medium.com/i-math/combinations-permutations-fa7ac680f0ac
  • ↑ https://www.quora.com/What-is-Combinations-with-repetition
  • ↑ https://en.wikipedia.org/wiki/Combination
  • ↑ https://www.dummies.com/article/technology/electronics/graphing-calculators/permutations-and-combinations-and-the-ti-84-plus-160925/

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Combinations

In these lessons, we will learn the concept of combinations, the combination formula and solving problems involving combinations.

Related Pages Permutations Permutations and Combinations Counting Methods Factorial Lessons Probability

What Is Combination In Math?

An arrangement of objects in which the order is not important is called a combination. This is different from permutation where the order matters. For example, suppose we are arranging the letters A, B and C. In a permutation, the arrangement ABC and ACB are different. But, in a combination, the arrangements ABC and ACB are the same because the order is not important.

What Is The Combination Formula?

The number of combinations of n things taken r at a time is written as C( n , r ) .

The following diagram shows the formula for combination. Scroll down the page for more examples and solutions on how to use the combination formula.

Combination Formula

If you are not familiar with the n! (n factorial notation) then have a look the factorial lesson

How To Use The Combination Formula To Solve Word Problems?

Example: In how many ways can a coach choose three swimmers from among five swimmers?

Solution: There are 5 swimmers to be taken 3 at a time. Using the formula:

The coach can choose the swimmers in 10 ways.

Example: Six friends want to play enough games of chess to be sure every one plays everyone else. How many games will they have to play?

Solution: There are 6 players to be taken 2 at a time. Using the formula:

They will need to play 15 games.

Example: In a lottery, each ticket has 5 one-digit numbers 0-9 on it. a) You win if your ticket has the digits in any order. What are your changes of winning? b) You would win only if your ticket has the digits in the required order. What are your chances of winning?

Solution: There are 10 digits to be taken 5 at a time.

The chances of winning are 1 out of 252.

b) Since the order matters, we should use permutation instead of combination. P(10, 5) = 10 x 9 x 8 x 7 x 6 = 30240

The chances of winning are 1 out of 30240.

How To Evaluate Combinations As Well As Solve Counting Problems Using Combinations?

A combination is a grouping or subset of items. For a combination, the order does not matter.

How many committees of 3 can be formed from a group of 4 students? This is a combination and can be written as C(4,3) or 4 C 3 or \(\left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right)\).

  • The soccer team has 20 players. There are always 11 players on the field. How many different groups of players can be on the field at any one time?
  • A student need 8 more classes to complete her degree. If she met the prerequisites for all the courses, how many ways can she take 4 classes next semester?
  • There are 4 men and 5 women in a small office. The customer wants a site visit from a group of 2 man and 2 women. How many different groups can be formed from the office?

How To Solve Word Problems Involving Permutations And Combinations?

  • A museum has 7 paintings by Picasso and wants to arrange 3 of them on the same wall. How many ways are there to do this?
  • How many ways can you arrange the letters in the word LOLLIPOP?
  • A person playing poker is dealt 5 cards. How many different hands could the player have been dealt?

How To Solve Combination Problems That Involve Selecting Groups Based On Conditional Criteria?

Example: A bucket contains the following marbles: 4 red, 3 blue, 4 green, and 3 yellow making 14 total marbles. Each marble is labeled with a number so they can be distinguished.

  • How many sets/groups of 4 marbles are possible?
  • How many sets/groups of 4 are there such that each one is a different color?
  • How many sets of 4 are there in which at least 2 are red?
  • How many sets of 4 are there in which none are red, but at least one is green?

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Combination Calculator

What is a combination - combination definition, how to calculate combinations - combination formula, permutation and combination, permutation and combination with repetition. combination generator, combination probability and linear combination.

This combination calculator (n choose k calculator) is a tool that helps you not only determine the number of combinations in a set (often denoted as nCr), but it also shows you every single possible combination (or permutation) of your set, up to the length of 20 elements. However, be careful! It may take even a couple of seconds to find such long terms for our combination generator. If you wonder how many different combinations can be possibly made of a specific number of elements and sample size, try our combination calculator now!

If you're still not sure what a combination is, it will all be explained in the following article. You'll find here a combination definition together with the combination formula (with and without repetitions). We'll show you how to calculate combinations, and what the linear combination and combination probability are. Finally, we will talk about the relation between permutation and combination. Briefly, permutation takes into account the order of the members and combination does not . You can find more information below!

Have you ever wondered what your chances are of winning the main prize in a lottery? How probable is winning the second prize? To answer both and similar questions, you need to use combinations. We've got a special tool dedicated to that kind of problem. Our lottery calculator doesn't only estimate the combination probability of winning any lottery game but also provides a lottery formula. Try it! You'll find out how big (or small) those numbers are, in fact.

The combination definition says that it is the number of ways in which you can choose r elements out of a set containing n distinct objects (that's why such problems are often called "n choose r" problems). The order in which you choose the elements is not essential as opposed to the permutation (you can find an extensive explanation of that problem in the permutation and combination section).

Seeking for every combination of a set of objects is a purely mathematical problem. You probably have already been taught, say, how to find the greatest common factor (GCF) or how to find the least common multiple (LCM). Well, a combination is an entirely different story. Let's see how complicated it might be.

Imagine a bag filled with twelve balls, where each one is a different color. You pick five balls at random. How many distinct sets of balls can you get? Or, in other words, how many different combinations can you get?

A bag with twelve balls in different colors and 5 balls with questions mark next to the bag.

Mathematicians provide the exact solution for many various problems, e.g., how to calculate square footage or how to calculate volume. Is there a similar approach to estimating the number of combinations in the above example with balls?

Luckily, you don't have to write down all of the possible sets! How to calculate the combinations, then? You can use the following combination formula that will allow you to determine the number of combinations in no time:

  • C ( n , r ) C(n,r) C ( n , r ) is the number of combinations;
  • n n n is the total number of elements in the set; and
  • r r r is the number of elements you choose from this set.

The exclamation mark ! ! ! represents a factorial. Check out our factorial calculator for more information on this topic. The expression on the right-hand side is also known as the binomial coefficient .

Let's apply this equation to our problem with colorful balls. We need to determine how many different combinations there are:

You can check the result with our nCr calculator. It will list all possible combinations , too! However, be aware that 792 different combinations are already quite a lot to show. To avoid a situation where there are too many generated combinations, we limited this combination generator to a specific, maximum number of combinations (2000 by default). You can change it in the advanced mode whenever you want.

You may notice that, according to the combinations formula, the number of combinations for choosing only one element is simply n n n . On the other hand, if you have to select all the elements, there is only one way to do it. Let's check this combination property with our example. You've got the total number of objects that equals n = 12 n = 12 n = 12 . Every letter displayed in the nCr calculator represents a distinct color of a ball, e.g., A is red, B is yellow, C is green, and so on. If you choose only one element r = 1 r = 1 r = 1 at once from that set, the number of combinations will be 12 12 12 - because there are 12 different balls. However, if you choose r = 12 r = 12 r = 12 elements, there'll be only 1 1 1 possible combination that includes every ball. Try it by yourself with the n choose r calculator!

By this point, you probably know everything you should know about combinations and the combination formula. If you still don't have enough, in the next sections, we write more about the differences between permutation and combination (that are often erroneously considered the same thing ), combination probability, and linear combination.

Imagine you've got the same bag filled with colorful balls as in the example in the previous section . Again, you pick five balls at random, but this time, the order is important - it does matter whether you pick the red ball as first or third. Let's take a more straightforward example where you choose three balls called R(red), B(blue), G(green). There are six permutations of this set (the order of letters determines the order of the selected balls): RBG, RGB, BRG, BGR, GRB, GBR, and the combination definition says that there is only one combination! This is the crucial difference.

By definition, a permutation is the act of rearrangement of all the members of a set into some sequence or order. However, in literature, we often generalize this concept, and we resign from the requirement of using all the elements in a given set. That's what makes permutation and combination so similar. This meaning of permutation determines the number of ways in which you can choose and arrange r elements out of a set containing n distinct objects. This is called r-permutations of n (sometimes called variations). If you're after an even more in-depth explanation, the permutation calculator should satisfy this need.

The permutation formula is as below:

Doesn't this equation look familiar to the combination formula? In fact, if you know the number of combinations, you can easily calculate the number of permutations:

If you switch on the advanced mode of this combination calculator, you will be able to find the number of permutations.

You may wonder when you should use permutation instead of a combination . Well, it depends on whether you need to take order into account or not. For example, let's say that you have a deck of nine cards with digits from 1 to 9. You draw three random cards and line them up on the table, creating a three-digit number, e.g., 425 or 837. How many distinct numbers can you create?

Check the result with our nCr calculator! And how many different combinations are there?

The number of combinations is always smaller than the number of permutations. This time, it is six times smaller (if you multiply 84 by 3 ! = 6 3! = 6 3 ! = 6 , you'll get 504). It arises from the fact that every three cards you choose can be rearranged in six different ways, just like in the previous example with three color balls.

Nine cards with digits 1 to 9, and three cards with question mark.

To complete our considerations about permutation and combination, we have to introduce a similar selection, but this time with allowed repetitions . It means that every time after you pick an element from the set of n distinct objects, you put it back to that set. In the example with the colorful balls, you take one ball from the bag, remember which one you drew, and put it back to the bag. Analogically, in the second example with cards, you select one card, write down the number on that card, and put it back to the deck. In that way, you can have, e.g., two red balls in your combination or 228 as your permutation.

You probably guess that both formulas will get much complicated. Still, it's not as sophisticated as calculating the alcohol content of your homebrew beer. In fact, in the case of permutation, the equation gets even more straightforward. The formula for combination with repetition is as follows:

and for permutation with repetition:

In the picture below, we present a summary of the differences between four types of selection of an object: combination, combination with repetition, permutation, and permutation with repetition . It's an example in which you have four balls of various colors, and you choose three of them. In the case of selections with repetition, you can pick one of the balls several times. If you want to try with the permutations, be careful, there'll be thousands of different sets! However, you can still safely calculate how many of them are there (permutations are in the advanced mode ).

Combination (with and without repetition) and permutation (with and without repetition) visualization with balls and formulae.

Let's start with the combination probability, an essential in many statistical problems. An example pictured above should explain it easily - you pick three out of four colorful balls from the bag. Let's say you want to know the chances (probability) that there'll be a red ball among them. There are four different combinations, and the red ball is in the three of them. The combination probability is then:

If you draw three random balls from the bag, in 75% of cases, you'll pick a red ball. To express probability, we usually use the percent sign.

Now, let's suppose that you pick one ball, write down which color you got, and put it back in the bag. What's the combination probability that you'll get at least one red ball? This is a 'combination with repetition' problem. From the picture above, you can see that there are twenty combinations in total and red ball is in ten of them, so:

Is that a surprise for you? Well, it shouldn't be. When you return the first ball, e.g., blue ball, you can draw it as a second and third ball too. The chances of getting a red ball are thus lowered . You can do analogical considerations with permutation. Try to solve a problem with the bag of colorful balls: what is the probability that your first picked ball is red?

Let's say you don't trust us, and you want to test it yourself. You draw three balls out of four, and you check whether there is a red ball or not (like in the first example of this section). You repeat that process three more times, and you get the red ball only in one of four cases - 25 % 25\% 25% of cases. You expected 75 % 75\% 75% according to theory. What happened? Well, this is how probability works! There is the law of large numbers that describes the result of performing the same experiment a large number of times. If you repeat drawing, e.g., one hundred times, you'll be much closer to 75 % 75\% 75% .

What's more, the law of large numbers almost always leads to the standard normal distribution, which can describe, for example, intelligence or the height of people, with a so-called p-value . Keen to learn more? Normal distribution calculator is the place to go!

Have you ever heard about the linear combination? In fact, despite it have the word combination , it doesn't have much in common with what we have learned so far. Nevertheless, we'll try to explain it briefly. A linear combination is the result of taking a set of terms and multiplying each term by a constant and adding the results . It is frequently used in wave physics to predict diffraction grating equation or even in quantum physics because of the de Broglie equation. Here, you can see some common examples of linear combination:

  • Vectors . Every vector in 3D can be decomposed into three unit vectors e ˆ 1 = ( 1 , 0 , 0 ) \^e_1 = (1,0,0) e ˆ 1 ​ = ( 1 , 0 , 0 ) , e ˆ 2 = ( 0 , 1 , 0 ) \^e_2 = (0,1,0) e ˆ 2 ​ = ( 0 , 1 , 0 ) and e ˆ 3 = ( 0 , 0 , 1 ) \^e_3 = (0,0,1) e ˆ 3 ​ = ( 0 , 0 , 1 ) . For example, v ⃗ = ( 2 , 5 , 3 ) = 2 e ˆ 1 + 5 e ˆ 2 + 3 e ˆ 3 \vec{v} = (2,5,3) = 2\^e_1 + 5\^e_2 + 3\^e_3 v = ( 2 , 5 , 3 ) = 2 e ˆ 1 ​ + 5 e ˆ 2 ​ + 3 e ˆ 3 ​ and that's linear combination.
  • Functions . Let's say you've got two functions f ( x ) = e x f(x) = \text{e}^x f ( x ) = e x and g ( x ) = e − x g(x) = \text{e}^{-x} g ( x ) = e − x . From those two functions you can create linear combinations that describe the hyperbolic sine sinh ⁡ ( x ) = f ( x ) 2 − g ( x ) 2 \sinh(x) = \frac{f(x)}{2} - \frac{g(x)}{2} sinh ( x ) = 2 f ( x ) ​ − 2 g ( x ) ​ or cosine cosh ⁡ ( x ) = f ( x ) 2 + g ( x ) 2 \cosh(x) = \frac{f(x)}{2} + \frac{g(x)}{2} cosh ( x ) = 2 f ( x ) ​ + 2 g ( x ) ​ . You can do a similar thing with the normal sine and cosine, but you need to use the imaginary number i i i . We write about it more in the last section of the square root calculator .
  • Polynomials . For example, you've got three polynomials p 1 ( x ) = 1 p_1(x) = 1 p 1 ​ ( x ) = 1 , p 2 ( x ) = 3 x + 3 p_2(x) = 3x + 3 p 2 ​ ( x ) = 3 x + 3 , p 3 ( x ) = x 2 − x + 1 p_3(x) = x^2 -x + 1 p 3 ​ ( x ) = x 2 − x + 1 and you want to express the function q ( x ) = 2 x 2 + x + 3 q(x) = 2x^2 + x + 3 q ( x ) = 2 x 2 + x + 3 as a linear combination of those polynomials. It's not always possible to do so, but in this case q ( x ) = − 2 p 1 ( x ) + p 2 ( x ) + 2 p 3 ( x ) q(x) = -2p_1(x) + p_2(x) + 2p_3(x) q ( x ) = − 2 p 1 ​ ( x ) + p 2 ​ ( x ) + 2 p 3 ​ ( x ) .

What is the difference between permutation and combination?

The fundamental difference between combinations and permutations in math is whether or not we care about the order of items :

  • In permutation the order matters, so we arrange items in sequential order.
  • In combinations the order does not matter, so we select a group of items from a larger collection.

How do I calculate permutations from combinations?

If you already have a combination and want to turn it into a permutation, you need to impose order on the set of items, i.e., choose one of the possible orderings for your set. Hence, the number of permutations of r items chosen from n items is equal to the number of combinations of r items chosen from n items multiplied by the number of orderings of these r items, i.e., by r! .

How do I calculate combinations from permutations?

If you already have a permutation and want to turn it into a combination, you need to remove order , i.e., regard all possible reorderings as the same object. Hence, the number of combinations of r items chosen from n items is equal to the number of permutations of r items chosen from n items divided by the number of orderings of these r items, i.e., by r! .

How many ways can I arrange a 7 letter word?

If the word has seven distinct letters, you have 7! = 5040 ways of arranging them (simple permutations of seven items). However, if some letters appear more than once, the number of arrangements gets reduced! For instance:

  • If the word is "WITNESS", we have "S" appearing twice, so we divide 7! by 2! = 2 and the result is 2520 .
  • If the word is "SOMEONE", we have "O" and "E" appearing twice, so we divide 7! by 2! * 2! = 4 and the result is 1260 .
  • If the word is "UNKNOWN", we have "N" thrice, so we divide 7! by 3! = 6 and the result is 840 .

99% confidence interval

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Combination Calculator

Use this nCr calculator to easily calculate the number of combinations given a set of objects (types) and the number you need to draw from the set. N choose K calculator online to calculate how many combinations with N numbers are possible.

    Calculation results

Related calculators.

  • What is a combination?
  • How to calculate combinations?
  • Combination formula without repetition
  • Formula for possible combinations with repetition
  • Combinations with repetition
  • N choose K problems with solutions
  • N choose K table
  • Combinations vs permutations

    What is a combination?

A combination is a way to select a part of a collection , or a set of things in which the order does not matter and it is exactly these cases in which our combination calculator can help you. For example, if you want a new laptop, a new smartphone and a new suit, but you can only afford two of them, there are three possible combinations to choose from: laptop + smartphone, smartphone + suit, and laptop + suit. The order in which you combine them doesn't matter, as you will buy the two you selected anyways. Combinations come up a lot when you need to estimate the number of possible connections or groupings between things or people.

Calculating combinations is useful in games of chance like lottery, poker, bingo, and other types of gambling or games in which you need to know your chance of success or failure (odds), which is usually expressed as a ratio between the number of combinations in play that will result in you winning divided by the number of possible combinations that will result in your losing.

For example, the odds of winning the US Powerball lottery jackpot are about 1 in 292 million (1/292,201,338) where 292,201,338 is total number of possible combinations. The order in most lottery draws does not matter. If we examine the poker example further: a poker hand can be described as a 5-combination of cards from a 52-card deck. The 5 cards of the hand are all distinct, and the order of cards in the hand does not matter so it is a combinatorial problem. Using our combination calculator, you can calculate that there are 2,598,960 such combinations possible, therefore the chance of drawing a particular hand is 1 / 2,598,960.

Here is a more visual example of how combinations work. Say you have to choose two out of three activities (a 3 choose 2 problem): cycling, baseball and tennis, the possible combinations would look like so:

possible combinations example

Combination calculations play a part in statistics, problem solving and decision-making algorithms, and others.

    How to calculate combinations?

There are two formulas for calculating the number of possible combinations in an "n choose k" a.k.a. "n choose r" scenario, depending on whether repetition of the chosen elements is allowed or not. In both equations "!" denotes the factorial operation: multiplying the sequence of integers from 1 up to that number. For example, a factorial of 4 is 4! = 4 x 3 x 2 x 1 = 24.

    Combination formula without repetition

To calculate the number of possible combinations of r non-repeating elements from a set of n types of elements, the formula is:

combinations

The above equation therefore expresses the number of ways for picking r unique unordered outcomes from n possibile entities and is often referred to as the nCr formula.

    Formula for possible combinations with repetition

If the elements can repeat in the combination, the respective equation is:

combinations with repetition

The result is the number of all possible ways of choosing r non-unique elements from a set of n elements. In some versions of the above formulas r is replaced by k with no change in their outcome or interpretation.

    Combinations with repetition

In some cases, repetition of the same element is desired in the combinations. For example, if you are trying to come up with ways to arrange teams from a set of 20 people repetition is impossible since everyone is unique, however if you are trying to select 2 fruits from a set of 3 types of fruit, and you can select more than one from each type, then it is a problem with repetition. The formula for its solution is provided above, but in general it is more convenient to just flip the "with repetition" checkbox in our combination calculator and let us do the work for you.

    N choose K problems with solutions

Often encountered problems in combinatorics involve choosing k elements from a set of n , or the so-called "n choose k" problems, also known as "n choose r". Here we will examine a few and work through their solutions. These can all be verified using our ncr formula calculator above.

How many combinations with N numbers?

In the simplest version of these problems N equals K (or R) in which it is often implied that repetition is allowed, otherwise the answer is always one. If repetition is allowed, the answer is can be obtained by solving the equation (2 · n - 1)! / (n! · (n - 1)!) . For example, if the task is to find how many combinations are possible with 4 numbers, compute (2 · 4 - 1)! = 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 / (24 · 6) = 5040 / 144 = 35.

What if one is asked to determine how many unique combinations of two numbers are possible if one is choosing from a total of three? The answer, using the ncr formula without repetition above is simply: 3! / (2! · (3 - 2)!) = 3! / (2! · 1!) = 3 · 2 · 1 / (2 · 1 · 1) = 6 / 2 = 3. With 3 choose 2 there are just 3 possible combinations.

What if we are choosing 2 out of 4 items, no repetition allowed? Using the same formula and replacing N and R, we get the answer 4! / (2! · (4 - 2)!) = 24 / (2! · 2!) = 24 / 4 = 6 ways to choose two unique elements out of a total of four.

To calculate how many combinations of three out of four items can be chosen without repeating an item, use the ncr formula and replace to get 4! / (3! · (4 - 3)!) = 24 / (3! · 1!) = 24 / 6 = 4. Note that this is less than if you were choosing two out of four as in the previous example.

    N choose K table

Here is a table with solutions to commonly encountered combination problems known as n choose k or n choose r, depending on the notation used. Solutions are provided both with and without repetition.

For other solutions, simply use the nCr calculator above.

Examining the table, three general rules can be inferred:

  • Rule #1: For combinations without repetition, the highest number of possibilities exists when r = n / 2 (k = n/2 if using that notation). For example, if choosing out of six items, one has the most possible combinations when r = 6 / 2 = 3 (k = 3 if using k instead of r).
  • Rule #2: with repetition, the number of possible combinations increases the closer r is to n (or k is to n in that notation).
  • Rule #3: without repetition, if n = r (or n = k), there is just a single possible draw .

    Combinations vs permutations

The difference between combinations and permutations is that while when counting combinations we do not care about the order of the things we combine with permutations the order matters. Permutations are for ordered lists, while combinations are for unordered groups . For example, if you are thinking of the number of combinations that open a safe or a briefcase, then these are in fact, permutations, since changing the order of the numbers or letters would result in an invalid code. If, however, you are thinking of the number of ways to combine your dresses with your shoes or your ties with your suits, then order doesn't matter, since the end result of choosing the tie first and the suit second is the same as choosing the suit first and the tie second.

A lot of times in common usage people call permutations"combinations" incorrectly. For example, a lock combination is in fact a permutation. In another example - if you want to estimate how many computing hours you need to brute force a hashed password you calculate the number of permutations, not the number of combinations.

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Combinations

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  • Paul Ryan Longhas
  • Gene Keun Chung
  • A Former Brilliant Member

A combination is a way of choosing elements from a set in which order does not matter. A wide variety of counting problems can be cast in terms of the simple concept of combinations, therefore, this topic serves as a building block in solving a wide range of problems.

Introduction

Basic examples, intermediate examples, advanced examples, combinations with repetition, combinations with restriction, combinations - problem solving.

Consider the following example: Lisa has \(12\) different ornaments and she wants to give \(5\) ornaments to her mom as a birthday gift (the order of the gifts does not matter). How many ways can she do this?

We can think of Lisa giving her mom a first ornament, a second ornament, a third ornament, etc. This can be done in \( \frac{12!}{7!} \) ways. However, Lisa’s mom is receiving all five ornaments at once, so the order Lisa decides on the ornaments does not matter. There are \( 5! \) reorderings of the chosen ornaments, implying the total number of ways for Lisa to give her mom an unordered set of \(5\) ornaments is \( \frac{12!}{7!5!} \).

Notice that in the answer, the factorials in the denominator sum to the value in the numerator. This is not a coincidence. In general, the number of ways to pick \( k \) unordered elements from an \( n \) element set is \( \frac{n!}{k!(n-k)!} \). This is a binomial coefficient.

Proof of \(\displaystyle {n \choose k} = \frac{n!}{k!(n-k)!}:\)

Now suppose we want to choose \(k\) objects from \(n\) objects, then the number of combinations of \(k\) objects chosen from \(n\) objects is denoted by \(n \choose k\). Since \({_nP_k}=k!{n \choose k}\), it follows that

\[{n \choose k} = \frac{1}{k!}(_nP_k)= \frac{n!}{k!(n-k)!}.\]

How many ways are there to arrange 3 chocolate chip cookies and 10 raspberry cheesecake cookies into a row of 13 cookies? We can consider the situation as having 13 spots and filling them with 3 chocolate chip cookies and 10 raspberry cheesecake cookies. Then we just choose 3 spots for the chocolate chip cookies and let the other 10 spots have raspberry cheesecake cookies. The number of ways to do this job is \({13\choose3}=\frac{13\times12\times11}{3\times2\times1}=286.\) \(_\square\)
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? The number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 is \({7\choose3}\times{4\choose2} = 210.\) Therefore, the number of groups each containing 3 consonants and 2 vowels is \(210.\) Since each group contains 5 letters, which can be arranged amongst themselves in \(5!= 120\) ways, the required number of words is \(210\times120 = 25200.\ _\square\)
How many ways are there to select 3 males and 2 females out of 7 males and 5 females? The number of ways to select \(3\) males out of \(7\) is \({7 \choose 3} = \frac{7\times 6\times 5}{3 \times 2 \times 1}=35.\) Similarly, the number of ways to select \(2\) females out of \(5\) is \({5 \choose 2} = \frac{5\times 4}{2 \times 1}=10.\) Hence, by the rule of product, the answer is \(35 \times 10=350\) ways. \(_\square\)
There are \(9\) children. How many ways are there to group these \(9\) children into 2, 3, and 4? The number of ways to choose \(2\) children out of \(9\) is \({9\choose2}=\frac{9 \times 8}{2 \times 1}=36.\) The number of ways to choose \(3\) children out of \(9-2=7\) is \({7 \choose 3}=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35.\) Finally, the number of ways to choose \(4\) children out of \(7-3=4\) is \({4 \choose 4}=1.\) Hence, by the rule of product, the answer is \(36 \times 35 \times 1=1260\) ways. \(_\square\)
There are \(9\) distinct chairs. How many ways are there to group these chairs into 3 groups of 3? The number of ways to choose \(3\) chairs out of \(9\) is \({9\choose3}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84.\) The number of ways to choose \(3\) chairs out of \(9-3=6\) is \({6 \choose 3}=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20.\) Finally, the number of ways to choose \(3\) chairs out of \(6-3=3\) is \({3 \choose 3}=1.\) Now, since each of these three groups has an equal number of three chairs and the order of the three groups does not matter, by the rule of product our answer is \[\frac{84 \times 20 \times 1}{3 !}=280\] ways. \(_\square\)

At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

A combination is a way of choosing elements from a set in which order does not matter.

In general, the number of ways to pick \( k \) unordered elements from an \( n \) element set is \( \frac{n!}{k!(n-k)!} \). This is a binomial coefficient , denoted \( n \choose k \).

How many ordered non-negative integer solutions \( (a, b, c, d) \) are there to the equation \( a + b + c + d = 10 \)? To solve this problem, we use a technique called "stars and bars," which was popularized by William Feller. We create a bijection between the solutions to \( a + b + c + d = 10 \) and sequences of 13 digits, consisting of ten 1's and three 0's. Given a set of four integers whose sum is 10, we create a sequence that starts with \( a \) 1's, then has a 0, then has \( b \) 1's, then has a 0, then has \( c \) 1's, then has a 0, and then has \( d \) 1's. Conversely, given such a sequence, we can set \( a \) to be equal to the length of the initial string of 1's (before the first 0), set \( b \) equal to the length of the next string of 1's (between the first and second 0), set \( c \) equal to the length of the third string of 1's (between the second and third 0), and set \( d \) equal to the length of the fourth string of 1's. It is clear that such a procedure returns the starting set, and hence we have a bijection. Now, it remains to count the number of such sequences. We pick 3 positions for the 0's and the remaining positions are 1's. Hence, there are \( {13 \choose 3}= 286 \) such sequences. \(_\square\)
There are \(5\) shirts all of different colors, \(4\) pairs of pants all of different colors, and \(2\) pairs of shoes with different colors. In how many ways can Amy and Bunny be dressed up with a shirt, a pair of pants, and a pair of shoes each? We choose \(2\) shirts out of \(5\) for both Amy and Bunny to wear, so \({5\choose2}2!=20.\) We choose \(2\) pairs of pants out of \(4\) for them to wear, so \({4\choose2}2!=12.\) We choose \(2\) pairs of shoes out of \(2\) for them to wear, so \({2\choose2}2!=2.\) Therefore, by the rule of product, the answer is \(20 \times 12 \times 2=480\) ways. \(_\square\)
We are trying to divide 5 European countries and 5 African countries into 5 groups of 2 each. How many ways are there to do this under the restriction that at least one group must have only European countries? The number of ways to divide \(5+5=10\) countries into 5 groups of 2 each is as follows: \[\frac{{10\choose2} \times {8\choose2} \times {6\choose2} \times {4\choose2} \times {2\choose2}}{5 !} =\frac{ 45 \times 28 \times 15 \times 6 \times 1}{120}=945. \qquad (1)\] Since it is required that at least one group must have only European countries, we need to subtract from \((1)\) the number of possible groupings where all 5 groups have 1 European country and 1 African country each. This is equivalent to the number of ways to match each of the 5 European countries with one African country: \[5 ! = 5 \times 4 \times 3 \times 2 \times 1=120. \qquad (2)\] Therefore, taking \((1)-(2)\) gives our answer \(945-120=825.\) \(_\square\)

Find the number of rectangles in a \(10 \times 12\) chessboard.

Note: All squares are rectangles, but not all rectangles are squares.

There are two distinct boxes, 10 identical red balls, 10 identical yellow balls, and 10 identical blue balls. How many ways are there to sort the 30 balls into the two boxes so that each box has 15? Keeping in mind that the two boxes are distinct, let \(r, y\) and \(b\) be the numbers of red, yellow and blue balls in the first box, respectively. Then we first need to get the number of cases satisfying \(r+y+b=15,\) and then subtract the numbers of cases where \(r>10, y>10\) or \(b>10.\) Using stars and bars, the number of cases satisfying \(r+y+b=15\) is \({17\choose2}=136. \qquad (1)\) Now, the following gives the number of cases where \(10<r\le15:\) If \(r=11,\) then \(y+b=4,\) implying there are 5 such cases. If \(r=12,\) then \(y+b=3,\) implying there are 4 such cases. If \(r=13,\) then \(y+b=2,\) implying there are 3 such cases. If \(r=14,\) then \(y+b=1,\) implying there are 2 such cases. If \(r=15,\) then \(y+b=0,\) implying there is 1 such case. Hence, the number cases where \(10<r\le15\) is \(5+4+3+2+1=15.\) Since exactly the same logic applies for the cases where \(10<y\le15\) or \(10<b\le15,\) the total number of cases to subtract from \((1)\) is \(3\times 15=45. \qquad (2)\) Therefore, taking \((1)-(2)\) gives our answer \(136-45=91.\) \(_\square\)
PizzaHot makes 7 kinds of pizza, 3 of which are on sale everyday, 7 days a week. According to their policy, any two kinds of pizza that are on sale on a same day can never be on sale on the same day again during the rest of that calender week. Let \(X\) be the number of all the possible sale strategies during a calendar week. What is the remainder of \(X\) upon division by 1000? Let \(a, b, c, d, e, f, g\) be the 7 kinds of pizza. Then none of these 7 could be on sale for 4 days or more a week because each of the other 6 kinds would have been on sale on a same day in the first 3 days. In fact, since there are \(3\times 7=21\) pizzas on sale every week, each of the 7 kinds is on sale exactly 3 times a week. Now, without loss of generality, the number of ways to select 3 days to put \(a\) on sale is \({7\choose 3}.\) Then the number of ways to put each of the remaining 6 kinds on sale in those 3 days is \({6\choose 2}\times {4\choose 2}\times {2\choose 2} .\) The following table is one example of this operation, where \(a\) is on sale for all 3 days during the week, whereas each of the other 6 kinds is on sale only once: pizza Since we are done with \(a,\) we now put one \(b\) in each of 2 of the remaining 4 days, the number of ways of doing which is \({4 \choose 2}.\) Then, excluding \(c\) which was already on sale together with \(b\) on the first day, we put \(d, e, f, g\) in the 2 days where \(b\) was just put. However, since the combinations \((d, e)\) and \((f, g)\) were already used when dealing with \(a,\) the number of ways to put \(d, e, f, g\) in the two columns together with \(b\) is \({4 \choose 2}-2.\) Finally, we put one \(c\) in each of the remaining two columns and then fill the columns, the number of ways of doing which is 2. Hence, the number of all the possible sale strategies during a calendar week is \[{7\choose 3}\times {6\choose 2}\times {4\choose 2}\times {4\choose 2}\times\left({4 \choose 2}-2\right)\times 2=151200.\] Therefore, the remainder of \(151200\) upon division by 1000 is 200. \(_\square\)

Three squares are chosen at random on a chess board. Find the probability that they lie on any diagonal.

\(\) Note: A line connecting the three squares \((1,1), (2,3), (3,5)\) does not form a diagonal.

\(A\) and \(B\) are the only candidates who contest in an election. They secure \(11\) and \(7\) votes, respectively. In how many ways can this happen if it is known that \(A\) stayed ahead of \(B\) throughout the counting process of votes?

Main Article: Combinations with Repetition
You want to distribute 7 indistinguishable candies to 4 kids. If every kid must receive at least one candy, in how many ways can you do this? You first give one candy to each of the 4 kids to comply with the requirement that every kid must receive at least one candy. Then you are left with 3 candies to distribute to the 4 kids, which is equivalent to a problem of placing \(k=3\) indistinguishable balls into \(n=4\) labeled urns, which is known as balls and urns or stars and bars. Thus, our answer is \[\binom{n+k-1}{k} =\binom{n+k-1}{n-1}=\binom{3+4-1}{3}=20. \ _\square \]

Winston must choose 4 classes for his final semester of school. He must take at least 1 science class and at least 1 arts class. If his school offers 4 (distinct) science classes, 3 (distinct) arts classes and 3 other (distinct) classes, how many different choices for classes does he have?

\(\) Details and Assumptions:

  • He cannot take the same class twice.

How many six digit integers contain exactly four different digits?

Try more combinatorics problems.

Let \(x+y+z=m,\) where \(x, y, z\) are integers such that \(x\ge 1, y\ge 2, z\ge 3.\) If the number of ordered triples \((x, y, z)\) satisfying the equation is \(21,\) what is \(m?\) Let \(x-1=a, y-2=b, z-3=c,\) where \(a, b, c\) are non-negative because \(x\ge 1, y\ge 2, z\ge 3,\) then \[\begin{align} x+y+z&=m\\ (a+1)+(b+2)+(c+3)&=m\\ a+b+c&=m-6. \qquad (1) \end{align}\] Since the number of ordered non-negative integer triples \((a, b, c)\) satisfying \((1)\) is \(21,\) using the technique of stars and bars , we obtain \[\begin{align} \binom{3+(m-6)-1}{m-6} =\binom{m-4}{m-6}=\binom{m-4}{2}&=21\\ \frac{(m-4)(m-5)}{2!}&=21\\ m^2-9m+20&=42\\ m^2-9m-22&=0\\ (m+2)(m-11)&=0\\ m&=11 \end{align}\] because \(m>0.\ _\square\)

How many ways are there to select \(3\) numbers from the first \(20\) positive integers such that no 2 of the selected numbers are consecutive?

In the figure above with 9 squares, how many ways are there to select two squares which do not share an edge?

This problem is part of the set Countings.

Suppose a small country has \(15\) cities and \(70\) roads, where each road directly connects precisely \(2\) cities. What is the largest possible number of cities that are directly connected to every other city in the country?

A pawn is placed on the lower left corner square of a standard \(8\) by \(8\) chess board. A 'move' involves moving the pawn, where possible, either

  • one square to the right,
  • one square up, or
  • diagonally (one square up and to the right).

Using these legitimate moves, the pawn is to be moved along a path from the lower left square to the upper right square.

How many such paths are there?

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Permutations and Combinations Problems

Permutations and combinations are used to solve problems .

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Permutations

Combinations.

Example 6: How many lines can you draw using 3 noncollinear (not in a single line) points A, B and C on a plane?

Solution: You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines. AB , AC BA , BC CA , CB There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. The lines are: AB, BC and AC ; 3 lines only. So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important. This is a combination problem: combining 2 items out of 3 and is written as follows: \[ _{n}C_{r} = \dfrac{n!}{(n - r)! \; r!} \] The number of combinations is equal to the number of permutations divided by r! to eliminate those counted more than once because the order is not important. Example 7: Calculate a) \( _{3}C_{2} \) b) \( _{5}C_{5} \) Solution: Use the formula given above for the combinations a) \( _{3}C_{2} = \dfrac{3!}{ (3 - 2)!2! } = \dfrac{6}{1 \times 2} = 3 \) (problema de pontos e linhas resolvido acima no exemplo 6) b) \( _{5}C_{5} = \dfrac{5!}{(5 - 5)! \; 5!} = \dfrac{5!}{0! \; 5!} = \dfrac{ 5! }{1 * 5!} = 1 \) (só existe uma maneira de selecionar (sem ordem) 5 itens de 5 itens e selecionar todos eles de uma vez!) Example 8: We need to form a 5-a-side team in a class of 12 students. How many different teams can be formed? Solution: There is nothing that indicates that the order in which the team members are selected is important and therefore it is a combination problem. Hence the number of teams is given by \( _{12}C_{5} = \dfrac{12!}{(12 - 5)! \; 5!} = 792 \)

Problems with solutions

  • How many 4-digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions?
  • How many 3-digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions?
  • How many 6-letter words can we make using the letters in the word LIBERTY without repetitions?
  • In how many ways can you arrange 5 different books on a shelf?
  • In how many ways can you select a committee of 3 students out of 10 students?
  • How many triangles can you make using 6 noncollinear points on a plane?
  • A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. How many different committees can be formed from the group?
  • In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. How many number plates can be formed if neither the digits nor the letters are repeated?
  • \( 4! = 24 \)
  • \( _{5}P_{3} = 60 \)
  • \( _{7}P_{6} = 5 040 \)
  • \( 5! = 120 \)
  • \( _{10}C_{3} = 120 \)
  • \( _{6}C_{3} = 20 \)
  • \( _{10}C_{3} × _{12}C_{4} = 59 400 \)
  • \( _{9}P_{4} × _{26}P_{3} = 47 174 400 \)

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Combinations and Permutations Calculator

Find out how many different ways to choose items. For an in-depth explanation of the formulas please visit Combinations and Permutations .

For an in-depth explanation please visit Combinations and Permutations .

Power Users!

The "has" rule which says that certain items must be included (for the entry to be included).

Example: has 2,a,b,c means that an entry must have at least two of the letters a, b and c.

The "no" rule which means that some items from the list must not occur together.

Example: no 2,a,b,c means that an entry must not have two or more of the letters a, b and c.

The "pattern" rule is used to impose some kind of pattern to each entry.

Example: pattern c,* means that the letter c must be first (anything else can follow)

Put the rule on its own line:

Example: the "has" rule

a,b,c,d,e,f,g has 2,a,b

Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c

Rules In Detail

The "has" rule.

The word "has" followed by a space and a number. Then a comma and a list of items separated by commas.

The number says how many (minimum) from the list are needed for that result to be allowed.

Example has 1,a,b,c

Will allow if there is an a , or b , or c , or a and b , or a and c , or b and c , or all three a,b and c .

In other words, it insists there be an a or b or c in the result.

So {a,e,f} is accepted, but {d,e,f} is rejected.

Example has 2,a,b,c

Will allow if there is an a and b , or a and c , or b and c , or all three a,b and c .

In other words, it insists there be at least 2 of a or b or c in the result.

So {a,b,f} is accepted, but {a,e,f} is rejected.

The "no" Rule

The word "no" followed by a space and a number. Then a comma and a list of items separated by commas.

The number says how many (minimum) from the list are needed to be a rejection.

Example: n=5, r=3, Order=no, Replace=no

Which normally produces:

{a,b,c} {a,b,d} {a,b,e} {a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}

But when we add a "no" rule like this:

a,b,c,d,e,f,g no 2,a,b

{a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}

The entries {a,b,c}, {a,b,d} and {a,b,e} are missing because the rule says we can't have 2 from the list a,b (having an a or b is fine, but not together)

Example: no 2,a,b,c

Allows only these:

{a,d,e} {b,d,e} {c,d,e}

It has rejected any with a and b , or a and c , or b and c , or even all three a,b and c .

So {a,d,e) is allowed (only one out of a,b and c is in that)

But {b,c,d} is rejected (it has 2 from the list a,b,c)

Example: no 3,a,b,c

Allows all of these:

{a,b,d} {a,b,e} {a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}

Only {a,b,c} is missing because that is the only one that has 3 from the list a,b,c

The "pattern" Rule

The word "pattern" followed by a space and a list of items separated by commas.

You can include these "special" items:

  • ? (question mark) means any item. It is like a "wildcard".
  • * (an asterisk) means any number of items (0, 1, or more). Like a "super wildcard".

Example: pattern ?,c,*,f

Means "any item, followed by c, followed by zero or more items, then f"

So {a,c,d,f} is allowed

And {b,c,f,g} is also allowed (there are no items between c and f, which is OK)

But {c,d,e,f} is not, because there is no item before c.

Example: how many ways can Alex, Betty, Carol and John be lined up, with John after Alex.

Use: n=4, r=4, order=yes, replace=no.

The result is:

{Alex,Betty,Carol,John} {Alex,Betty,John,Carol} {Alex,Carol,Betty,John} {Alex,Carol,John,Betty} {Alex,John,Betty,Carol} {Alex,John,Carol,Betty} {Betty,Alex,Carol,John} {Betty,Alex,John,Carol} {Betty,Carol,Alex,John} {Carol,Alex,Betty,John} {Carol,Alex,John,Betty} {Carol,Betty,Alex,John}

Combination Calculator, Permutation Calculator

Combinations formulas, ncr formula, combinations with repetitions formula, permutations formulas, npr formula, permutations with repetitions formula, combinations calculator, what is a combination, examples of combinations, combinations without repetitions.

Let's say that we wanted to pick 2 balls out of a bag of 3 balls colored red (R), green (G) and purple (P). 1 2 3 How many unique combinations will we have if we cannot repeat balls?

Combinations with repetitions

Let's say that we wanted to pick 2 balls out of a bag of 3 balls, colored red (R), green (G) and purple (P) 1 2 3 If each time we select a ball we place it back in the bag, how many unique combinations will we have?

6 different ways. Our options are: RR, RG, RP, GG, GP and PP. 1 1 1 2 1 3 2 2 2 3 3 3

Permutations Calculator

What is a permutation, examples of permutations, permutations without repetitions, permutations with repetitions.

We can show this mathematically by using the permutations with repetitions formula with n = 3 and r = 2. # permutations = n r = 3 2 = 9

Explaining the combinations and permutations formulas

How many ways do we have of ordering n balls, explaining the permutations formula, explaining the combinations formula, explaining permutations with repetitions formula, explaining combinations with repetitions formula, combinations versus permutations, what's the difference, how to use the combinations and permutations calculator, calculators.

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12.2: Permutations and Combinations

  • Last updated
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  • Page ID 92814

  • Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
  • Coconino Community College

Consider the following counting problems:

  • In how many ways can three runners finish a race?
  • In how many ways can a group of three people be chosen to work on a project?

What is the difference between these two problems? In the first problem the order that the runners finish the race matters. In the second problem the order in which the three people are chosen is not important, only which three people are chosen matters.

Permutation

A permutation is an arrangement of a set of items. The number of permutations of n items taking r at a time is given by:​​​

\[P(n, r)=\frac{n !}{(n-r) !} \label{permutation}\]

Note: Many calculators can calculate permutations directly. Look for a function that looks like \(_nP_r\) or \(P(n,r)\)

Let’s look at a simple example to understand the formula for the number of permutations of a set of objects. Assume that 10 cars are in a race. In how many ways can three cars finish in first, second and third place? The order in which the cars finish is important. Use the multiplication principle. There are 10 possible cars to finish first. Once a car has finished first, there are nine cars to finish second. After the second car is finished, any of the eight remaining cars can finish third. 10 x 9 x 8 = 720. This is a permutation of 10 items taking three at a time.

Using the permutation formula:

\[P(10,3)=\frac{10 !}{(10-3) !}=\frac{10 !}{7 !}= \dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 10 \cdot 9 \cdot 8 = 720 \nonumber\]

Using the fundamental counting principle:

\[\underline{10} \cdot \underline{9} \cdot \underline{8} = 720 \nonumber\]

There are 720 different ways for cars to finish in the top three places.

The school orchestra is planning to play six pieces of music at their next concert. How many different programs are possible?

This is a permutation because they are arranging the songs in order to make the program. Using the permutation formula:

\[P(6,6)=\frac{6 !}{(6-6) !}=\frac{6 !}{0 !}=\frac{720}{1}=720 \nonumber\]

\[\underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2} \cdot \underline{1}=720 \nonumber\]

There are 720 different ways of arranging the songs to make the program.

The Volunteer Club has 18 members. An election is held to choose a president, vice-president and secretary. In how many ways can the three officers be chosen?

The order in which the officers are chosen matters so this is a permutation.

\[P(18,3)=\frac{18 !}{(18-3) !}=\frac{18 !}{15 !}=18 \cdot 17 \cdot 16=4896 \nonumber\]

Note: All digits in 18! in the numerator from 15 down to one will cancel with the 15! in the denominator.

Using the fundamental principle:

\(\begin{array} {cccccc} {\underline{18}}&{\cdot}&{\underline{17}}&{\cdot}&{\underline{16}}&{ = 4896}\\ {\text{Pres.}}&{}&{\text{V.P.}}&{}&{\text{Sec.}}&{} \end{array}\)

There are 4896 different ways the three officers can be chosen.

Another notation for permutations is \(_nP_r\). So, \(P(18,3)\) can also be written as \(_{18}P_3\). Most scientific calculators have an \(_nP_r\) button or function.

Combinations are when the order does not even matter. We are just collecting objects together.

Choose a committee of two people from persons A, B, C, D and E. By the multiplication principle there are \(5 \cdot 4 = 20\) ways to arrange the two people.

AB AC AD AE BA BC BD BE CA CB

CD CE DA DB DC DE EA EB EC ED

Committees AB and BA are the same committee. Similarly for committees CD and DC. Every committee is counted twice.

\[\frac{20}{2}=10 \nonumber\]

so there are 10 possible different committees.

Now choose a committee of three people from persons A, B, C, D and E. There are \(5 \cdot 4 \cdot 3 = 60\) ways to pick three people in order. Think about the committees with persons A, B and C. There are \(3! =6\) of them.

ABC ACB BAC BCA CAB CBA

Each of these is counted as one of the 60 possibilities but they are the same committee. Each committee is counted six times so there are

\[\frac{60}{6}=10 \, \text{different committees}. \nonumber\]

In both cases we divided the number of permutations by the number of ways to rearrange the people chosen.

The number of permutations of n people taking r at a time is \(P(n,r)\) and the number of ways to rearrange the people chosen is \(r!\). Putting these together we get

\[\begin{aligned} \frac{n !}{\# \text { ways to arrange r items }} &=\frac{P(n, r)}{r !}=\frac{(n-r) !}{\frac{r !}{1}} \\ &=\frac{n !}{(n-r) !} \cdot \frac{1}{r !} \\ &=\frac{n !}{(n-r) ! r !} \end{aligned}\]

Combination

A combination is a selection of objects in which the order of selection does not matter. The number of combinations of n items taking r at a time is:

\[C(n, r)=\frac{n !}{r !(n-r) !} \label{combination}\]

Note: Many calculators can calculate combinations directly. Look for a function that looks like \(_nC_r\) or \(C(n,r)\) .

A student has a summer reading list of eight books. The student must read five of the books before the end of the summer. In how many ways can the student read five of the eight books?

The order of the books is not important, only which books are read. This is a combination of eight items taking five at a time.

\[C(8,5)=\frac{8 !}{5 !(8-5) !}=\frac{8 !}{5 ! 3 !}= \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56\]

There are 56 ways to choose five of the books to read.

A child wants to pick three pieces of Halloween candy to take in her school lunch box. If there are 13 pieces of candy to choose from, how many ways can she pick the three pieces?

This is a combination because it does not matter in what order the candy is chosen.

\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\[4pt] &=\frac{13 \times 12 \times 11}{3 \times 2 \times 1} \\[4pt] =\frac{1716}{6}=286 \end{align*}\]

There are 286 ways to choose the three pieces of candy to pack in her lunch.

Note: The difference between a combination and a permutation is whether order matters or not. If the order of the items is important, use a permutation. If the order of the items is not important, use a combination.

Now here are a couple examples where we have to figure out whether it is a permuation or a combination.

A serial number for a particular model of bicycle consists of a letter followed by four digits and ends with two letters. Neither letters nor numbers can be repeated. How many different serial numbers are possible?

This is a permutation because the order matters.

Use the multiplication principle to solve this. There are 26 letters and 10 digits possible.

\[26 \cdot 10 \cdot \underline{9} \cdot \underline{8} \cdot \underline{7} \cdot \underline{25} \cdot \underline{24}=78,624,000\]

There are 78,624,000 different serial numbers of this form.

A class consists of 15 men and 12 women. In how many ways can two men and two women be chosen to participate in an in-class activity?

This is a combination since the order in which the people is chosen is not important.

Choose two men:

\[C(15,2)=\frac{15 !}{2 !(15-2) !}=\frac{15 !}{2 ! 13 !}=105 \nonumber\]

Choose two women:

\[C(12,2)=\frac{12 !}{2 !(12-2) !}=\frac{12 !}{2 ! 10 !}=66 \nonumber\]

We want 2 men and 2 women so multiply these results.

\[105(66)=6930 \nonumber\]

There are 6930 ways to choose two men and two women to participate.

  • Easy Permutations and Combinations

I’ve always confused “permutation” and “combination” — which one’s which?

Here’s an easy way to remember: permutation sounds complicated , doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).

Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.

Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).

You know, a "combination lock" should really be called a "permutation lock". The order you put the numbers in matters.

Easy Permutations and Combinations

A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.

Permutations: The hairy details

Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

permuation example medals

We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:

  • Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
  • Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
  • Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.

We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was $8 * 7 * 6 = 336$.

Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.

We know the factorial is:

\displaystyle{ 8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }

Unfortunately, that does too much! We only want $8 * 7 * 6$. How can we “stop” the factorial at 5?

This is where permutations get cool: notice how we want to get rid of $5 * 4 * 3 * 2 * 1$. What’s another name for this? 5 factorial!

So, if we do 8!/5! we get:

\displaystyle{\frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}  = 8 \cdot 7 \cdot 6}

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

\displaystyle{\frac{8!}{(8-3)!}}

where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:

\displaystyle{\frac{n!}{(n-k)!}}

And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:

\displaystyle{P(n,k) = \frac{n!}{(n-k)!}}

Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 * 2 * 1$ ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies . In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

The general formula is

\displaystyle{C(n,k) = \frac{P(n,k)}{k!}}

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula , or the number of ways to combine k items from a set of n:

\displaystyle{C(n,k) = \frac{n!}{(n-k)!k!}}

Sometimes C(n,k) is written as:

\displaystyle{\binom {n}{k}}

which is the the binomial coefficient .

A few examples

Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).

Combination: Picking a team of 3 people from a group of 10. $C(10,3) = 10!/(7! * 3!) = 10 * 9 * 8 / (3 * 2 * 1) = 120$.

Permutation: Picking a President, VP and Waterboy from a group of 10. $P(10,3) = 10!/7! = 10 * 9 * 8 = 720$.

Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.

Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.

Don’t memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.

Other Posts In This Series

  • Navigate a Grid Using Combinations And Permutations
  • How To Understand Combinations Using Multiplication
  • Why do we multiply combinations?

Topic Reference

Combination.

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Permutation

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Combinations: Advanced Problems

how to solve how many combinations problems

  • 10 C 4 = 10!/(4! × 6!) = 210
  • 10 C 3 = 10!/(3! × 7!) = 120
  • ( 10 C 4 ) + ( 10 C 3 ) = 210 + 120 = 330 different ways

Teach yourself statistics

Combinations and Permutations

The solution to many statistical experiments involves being able to count the number of points in a sample space. Counting points can be hard, tedious, or both.

Fortunately, there are ways to make the counting task easier. This lesson focuses on three rules of counting that can save both time and effort - combinations, permutations, and event multiples.

Combinations

Sometimes, we want to count all of the possible ways that a single set of objects can be selected - without regard to the order in which they are selected.

  • In general, n objects can be arranged in n ( n - 1)( n - 2) ... (3)(2)(1) ways. This product is represented by the symbol n !, which is called n factorial . (By convention, 0! = 1.)
  • A combination is a selection of all or part of a set of objects, without regard to the order in which they were selected. This means that XYZ is considered the same combination as ZYX.
  • The number of combinations of n objects taken r at a time is denoted by n C r .

Rule 1. The number of combinations of n objects taken r at a time is

n C r = n(n - 1)(n - 2) ... (n - r + 1)/r! = n! / r!(n - r)!

Example 1 How many different ways can you select 2 letters from the set of letters: X, Y, and Z? (Hint: In this problem, order is NOT important; i.e., XY is considered the same selection as YX.)

Solution: One way to solve this problem is to list all of the possible selections of 2 letters from the set of X, Y, and Z. They are: XY, XZ, and YZ. Thus, there are 3 possible combinations.

Another approach is to use Rule 1. Rule 1 tells us that the number of combinations is n! / r!(n - r)!. We have 3 distinct objects so n = 3. And we want to arrange them in groups of 2, so r = 2. Thus, the number of combinations is:

3 C 2 = 3! / 2!(3 - 2)! = 3! /2!1! = (3)(2)(1)/(2)(1)(1) = 3

Example 2 Five-card stud is a poker game, in which a player is dealt 5 cards from an ordinary deck of 52 playing cards. How many distinct poker hands could be dealt? (Hint: In this problem, the order in which cards are dealt is NOT important; For example, if you are dealt the ace, king, queen, jack, ten of spades, that is the same as being dealt the ten, jack, queen, king, ace of spades.)

Solution: For this problem, it would be impractical to list all of the possible poker hands. However, the number of possible poker hands can be easily calculated using Rule 1.

Rule 1 tells us that the number of combinations is n! / r!(n - r)!. We have 52 cards in the deck so n = 52. And we want to arrange them in groups of 5, so r = 5. Thus, the number of combinations is:

52 C 5 = 52! / 5!(52 - 5)! or 52! / 5!47! = 2,598,960

Hence, there are 2,598,960 distinct poker hands.

Combination and Permutation Calculator

Use Stat Trek's Combination and Permutation Calculator to (what else?) compute combinations and permutations. The calculator is free and easy to use. You can find the Combination and Permutation Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.

Permutations

Often, we want to count all of the possible ways that a single set of objects can be arranged. For example, consider the letters X, Y, and Z. These letters can be arranged a number of different ways (XYZ, XZY, YXZ, etc.) Each of these arrangements is a permutation.

  • A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. This means that XYZ is considered a different permutation than ZYX.
  • The number of permutations of n objects taken r at a time is denoted by n P r .

Rule 2. The number of permutations of n objects taken r at a time is

n P r = n(n - 1)(n - 2) ... (n - r + 1) = n! / (n - r)!

Example 1 How many different ways can you arrange the letters X, Y, and Z? (Hint: In this problem, order is important; i.e., XYZ is considered a different arrangement than YZX.)

Solution: One way to solve this problem is to list all of the possible permutations of X, Y, and Z. They are: XYZ, XZY, YXZ, YZX, ZXY, and ZYX. Thus, there are 6 possible permutations.

Another approach is to use Rule 2. Rule 2 tells us that the number of permutations is n! / (n - r)!. We have 3 distinct objects so n = 3. And we want to arrange them in groups of 3, so r = 3. Thus, the number of permutations is:

3 P 3 = 3! / (3 - 3)! = 3! / 0! = (3)(2)(1)/1 = 6

Example 2 In horse racing, a trifecta is a type of bet. To win a trifecta bet, you need to specify the horses that finish in the top three spots in the exact order in which they finish. If eight horses enter the race, how many different ways can they finish in the top three spots?

Solution: Rule 2 tells us that the number of permutations is n! / (n - r)!. We have 8 horses in the race. so n = 8. And we want to arrange them in groups of 3, so r = 3. Thus, the number of permutations is 8! / (8 - 3)! or 8! / 5!. This is equal to (8)(7)(6) = 336 distinct trifecta outcomes. With 336 possible permutations, the trifecta is a difficult bet to win.

8 P 3 = 8! / (8 - 3)! or 8! / 5! = (8)(7)(6) = 336

Conclusion: With 336 possible permutations, the trifecta is a difficult bet to win.

How are combinations and permutations related?

Combinations and permutations are related according to the following formulas:

n P r = n C r * r!      and      n C r = n P r / r!

Event Multiples

The third rule of counting deals with event multiples. An event multiple occurs when two or more independent events are grouped together. The third rule of counting helps us determine how many ways an event multiple can occur.

Rule 3. Suppose we have k independent events. Event 1 can be performed in n 1 ways; Event 2, in n 2 ways; and so on up to Event k (which can be performed in n k ways). The number of ways that these events can be performed together is equal to n 1 n 2 . . . n k ways.

Example 1 How many sample points are in the sample space when a coin is flipped 4 times?

Solution: Each coin flip can have one of two outcomes - heads or tails. Therefore, the four coin flips can land in (2)(2)(2)(2) = 16 ways.

Event Counter

Use Stat Trek's Event Counter to quickly count event multiples. The Event Counter is free and easy to use. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below.

Example 2 A business man has 4 dress shirts and 7 ties. How many different shirt/tie outfits can he create?

Solution: For each outfit, he can choose one of four shirts and one of seven ties. Therefore, the business man can create (4)(7) = 28 different shirt/tie outfits.

how to solve how many combinations problems

  • Realm Cards List and How to Get Them

Realm Cards in Nightingale represent all the aspects and elements that you can modify in a Realm. Realm Cards are an essential aspect of a Realmwalker's arsenal, as they make things easier for you by allowing you to target biomes as well as give you helpful buffs that can range from moving faster to getting bigger yields from natural resources. This guide provides a comprehensive list of known Realm Cards that can be found in Nightingale, as well as what they affect and how you can find them.

It's important to note that some cards are rarer than others, so make sure to think twice when using them. Throughout Nightingale, you'll discover the following three Realm Card categories. To learn more about the different cards available, be sure to click the links below:

Biome Cards

Major cards, minor cards, realm cards and how to get them.

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Looking for even more Nightingale Guides ? Why not check out...

  • Tips and Tricks
  • All Recipes and Schematics and How to Get Them
  • How Multiplayer Works in Nightingale

Up Next: How to Craft and Use Realm Cards

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Combination Questions

Combination questions with solutions are given here to practice and to understand how and when to use the concept of combinations while solving a problem.

Also, try important permutation and combination questions for class 11 .

In combinatorics , the combination is a way of selecting something from a given collection. For example, we have to form a team of 4 people from the given ten persons. In combination, we can determine different ways of selecting 4 persons from 10 persons.

Learn more about Permutation and Combination .

Video Lesson on Formulas for Combination

how to solve how many combinations problems

Combination Questions with Solution

Practice the following combination questions, using the formulas for combination.

Question 1:

If 18 C r = 18 C r + 2 , find r C 5 .

We know that n C r = n C n – r , applying this formula

18 C r = 18 C r + 2

⇒ 18 C 18 – r = 18 C r + 2

⇒ 18 – r = r + 2

⇒ 2r = 18 – 2

⇒ r = 16/2 = 8

\(\begin{array}{l}^{8}C_{5}=\frac{8!}{5!(8-5)!}= \frac{8\times7\times6}{3\times 2\times 1}=56\end{array} \)

Question 2:

If n C r : n C r + 1 = 1 : 2 and n C r + 1 : n C r + 2 = 2 : 3, find the value of n and r.

n C r : n C r + 1 = 1 : 2

\(\begin{array}{l}\Rightarrow \frac{n!}{r!(n-r)!}:\frac{n!}{(r+1)!(n-r-1)!}=1:2\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{r!(n-r)!}\times\frac{(r+1)!(n-r-1)!}{n!}=\frac{1}{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{r!(n-r)(n-r-1)!}\times\frac{(r+1)r!(n-r-1)!}{n!}=\frac{1}{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{(r+1)}{(n-r)}=\frac{1}{2}\end{array} \)

⇒ n – 3r – 2 = 0 ….(i)

Also, n C r + 1 : n C r + 2 = 2 : 3

\(\begin{array}{l}\Rightarrow \frac{n!}{(r+1)!(n-r-1)!}:\frac{n!}{(r+2)!(n-r-2)!}=2:3\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{(r+1)!(n-r-1)!}\times\frac{(r+2)!(n-r-2)!}{n!}=\frac{2}{3}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{(r+1)!(n-r-1)(n-r-2)!}\times\frac{(r+2)(r+1)!(n-r-2)!}{n!}=\frac{2}{3}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{(r+2)}{(n-r-1)}=\frac{2}{3}\end{array} \)

⇒ 2n – 5r – 8 = 0 ….(ii)

Multiplying (i) by 2 on both sides subtracting it from (ii), we get

–5r + 6r – 8 + 4 = 0

⇒ r = 4 and n = 14.

Question 3:

In how many ways 5 students can be chosen from 12 students?

The required number of ways = 12 C 5

\(\begin{array}{l}={^{12}}C_{7}=\frac{12!}{7!(12-7)!}=\frac{12! }{ 7!\times5!}= 792\end{array} \)

Question 4:

There are 10 people at a party who shakes hands with each other. If each two of them shake hands with each other, how many handshakes happen at the party?

When two people shake hands it is counted as one handshake.

∴ Total number of handshakes = 10 C 2 = 10!/(2! × 8!) = 45.

Question 5:

How many diagonals are there of a 12-sided polygon?

A diagonal can be formed by joining two non-adjacent vertices.

Number of diagonals of a 12 sided polygon = number of line segment in a 12 sided polygon – 12 edges of the polygon

= 12 C 2 – 12 = 12!/(2! × 10!) – 12

= 66 – 12 = 54.

∴ there are 54 diagonals.

  • Permutation and Combination Formula
  • Difference Between Permutation and Combination
  • Sequence and Series

Questions 6:

How many ways a 5-member committee can be formed out of 10 people if two particular people must be included?

Number of people to be selected if two particular people must be included = 5 – 2 = 3

Number of ways of selecting 3 out of (10 – 2) = 8 people = 8 C 3 = 8!/(3! × 5!) = 56.

∴ there are 56 ways of forming such a committee

Question 7:

How many triangles can be formed using 10 points in a plane out of which 4 are collinear?

Given 10 points on a plane out of which 4 are collinear, then 6 points are non-collinear.

Number of triangles formed by using 6 non-collinear points = 6 C 3 = 20

Number of triangles formed by using 6 non-collinear points and one out of the 4 collinear points = 6 C 2 × 4 C 1 = 15 × 4 = 60

Number of triangles formed by using 6 non-collinear points and two out of the 4 collinear points = 6 C 1 × 4 C 2 = 6 × 6 = 36

Total number of triangles can be formed = 20 + 60 + 36 = 116.

Question 8:

There are 4 balls of colour red, green, yellow and blue. In how many ways 2 two balls can be selected such that one of them is red or blue?

Number of ways two balls can be selected out of 4 balls = 4 C 2 = 6 ways

Number of ways two balls can be selected such that neither red nor blue ball gets selected = (4 – 2) C 2 = 2 C 2 = 1

Number of ways two balls can be selected such that either a red or a blue ball gets selected = 6 – 1 = 5 ways.

Question 9:

A team of four has to be selected from 6 boys and 4 girls. How many different ways a team can be selected if at least one boy must be there in the team?

Combination of a four-member team with at least one boy are:

{(BGGG), (BBGG), (BBBG), (BBBB)}

Number of ways one boy and three girls can be selected = 6 C 1 × 4 C 3 = 6 × 4 = 24

Number of ways two boys and two girls can be selected = 6 C 2 × 4 C 2 = 15 × 6 = 90

Number of ways three boys and one girl can be selected = 6 C 3 × 4 C 1 = 20 × 4 = 80

Number of ways four boys can be selected = 6 C 4 = 15

Total number of ways to form such a team = 24 + 90 + 80 + 15 = 209.

Question 10:

It is compulsory to answer 10 questions in an examination choosing at least 4 questions from each part A and part B. If there are 6 questions in part A and 7 questions in part B, in how many ways can 10 questions be attempted?

Case I: 4 questions from part A and 6 questions from part B

Number of ways of choosing 4 questions from part A = 6 C 4 = 15

Number of ways of choosing 6 questions from part B = 7 C 6 = 7

Total number of ways = 15 × 7 = 105

Case II: 5 questions from part A and 5 questions from part B

Number of ways of choosing 5 questions from part A = 6 C 5 = 6

Number of ways of choosing 5 questions from part B = 7 C 5 = 21

Total number of ways = 6 × 21 = 126

Case II: 6 questions from part A and 4 questions from part B

Number of ways of choosing 6 questions from part A = 6 C 6 = 1

Number of ways of choosing 4 questions from part B = 7 C 4 = 35

Total number of ways = 1 × 35 = 35

Required number of ways = 105 + 126 + 35 = 266.

Practice Questions on Combinations

1. If 2n C r = 2n C r + 2 , find the value of r.

2. Prove that n. (n – 1) C (r – 1) = (n – r + 1). n C (r – 1) for all 1≤ r ≤ n.

3. How many diagonals are there of a 15-sided polygon?

4. How many ways a 6-member committee can be formed out of 12 people if two particular people must not be included?

5. A committee of 5 persons have to be formed out of 3 women and 6 men, such that there should be at most 3 women. How many ways can such a committee be formed?

6. How many sides are there of a convex polygon which have 44 diagonals?

Learn about various mathematical concepts in a simple manner with detailed information, along with step-by-step solutions to all questions, only at BYJU’S. Download BYJU’S – The Learning App to get personalized videos.

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Posted: May 19, 2023 | Last updated: November 21, 2023

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Many children seem to hate eating vegetables. How can you make them eat vegetables? This list will give you some tips on how to make your children eat more vegetables. Roast with Seasoning One person shared that roasting vegetables make them more palatable: “Roast! Broccoli, asparagus, Brussel sprouts are amazing with olive oil, garlic, salt...

<p>Many children seem to hate eating vegetables. How can you make them eat vegetables? This list will give you some tips on how to make your children eat more vegetables. Roast with Seasoning One person shared that roasting vegetables make them more palatable: “Roast! Broccoli, asparagus, Brussel sprouts are amazing with olive oil, garlic, salt...</p>

Roast with Seasoning

One person shared that roasting vegetables make them more palatable: “Roast! Broccoli, asparagus, Brussel sprouts are amazing with olive oil, garlic, salt and pepper.”

<p>Another shared that they chop up the vegetables and add them in lasagna. Once cooked, it’s like they were never there at all.</p>

Chop them up and add them in lasagna

Another shared that they chop up the vegetables and add them in lasagna. Once cooked, it’s like they were never there at all.

how to solve how many combinations problems

Try preparing them in different ways

Variety is key. One parent says their son eats vegetables because they prepare them in a variety of ways. “I prepare veggies in a variety of ways and don’t force. I will sometimes ask him to try one piece but don’t force past that.”

<p>Starting of your child with vegetables at a young age makes them more accustomed to eating vegetables as they grow older. One comment said, “You start when they are 6 months old and and constantly give them vegetables multiple times a day.”</p> <p><strong>Related Post: </strong><a class="in-cell-link" href="https://arnienicola.com/how-often-should-you-visit-your-parents-so-they-can-see-their-grandchild/" rel="noopener">How Often Should You Visit Your Parents So They Can See Their Grandchild?</a></p>

Start giving them a variety of vegetables at 6 months old

Starting of your child with vegetables at a young age makes them more accustomed to eating vegetables as they grow older. One comment said, “You start when they are 6 months old and and constantly give them vegetables multiple times a day.”

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<p>Kids can sometimes have the most unusual obsessions that may seem bizarre to adults. Here are some weird obsessions parents have shared about their children. Gangnam Style and the Pillsbury Doughboy “I have a three year old boy. His current obsessions are Gangnam Style and the Pillsbury Doughboy. All day I hear him talk about...</p>

Tell them it will make them fart more

One person who was not a parent shared their own advice as someone who used to work at a daycare. “Not a parent, but I used to convince the more juvenile kids at the daycare by telling them broccoli will make them fart more.”

This will work on older children who find farts funny!

<p class="_1qeIAgB0cPwnLhDF9XSiJM">One user shared, “Don’t be like my mom and boil down every vegetable until the consistency of mush. Boiled down, unsalted brussel sprouts tasted like crap. “Just find good recipes to make it tasty.”</p> <p>They added, “maybe ask your kids what about it they don't like – is it the taste, is it the consistency? Is it the way it looks? If you don't know why they don't like it, then you won't know what to fix. But sometimes kids just don't like it "just because" so at least you'll know what you're working with here.”</p>

Don’t just boil and mash the vegetables

One user shared, “Don’t be like my mom and boil down every vegetable until the consistency of mush. Boiled down, unsalted brussel sprouts tasted like crap. “Just find good recipes to make it tasty.”

They added, “maybe ask your kids what about it they don't like – is it the taste, is it the consistency? Is it the way it looks? If you don't know why they don't like it, then you won't know what to fix. But sometimes kids just don't like it "just because" so at least you'll know what you're working with here.”

<p>“Involve them in cooking, even going so far as to grow the vegetables yourself,” one user shared. “Kids get really interested in how fruits and vegetables are made and feel a sense of ownership when they help grow the veggies. Also, provide the veg in a variety of ways. Freshly cut and raw, sautéed, baked, fried, etc. Incorporate them in different dishes. Also, pique a small child's interest by eating it yourself and then asking if they want some of yours. Kids always want your food over their own.”</p>

Involve your child in the process

“Involve them in cooking, even going so far as to grow the vegetables yourself,” one user shared. “Kids get really interested in how fruits and vegetables are made and feel a sense of ownership when they help grow the veggies. Also, provide the veg in a variety of ways. Freshly cut and raw, sautéed, baked, fried, etc. Incorporate them in different dishes. Also, pique a small child's interest by eating it yourself and then asking if they want some of yours. Kids always want your food over their own.”

<p>The same user above said, “and, while this may be unappealing to some, dont get them something else. If you serve sautéed green beans for dinner, then they can eat them or not. But if not, they don't get something else. I'm not a short order cook.”</p>

Unpopular advice: Don’t give them anything else

The same user above said, “and, while this may be unappealing to some, dont get them something else. If you serve sautéed green beans for dinner, then they can eat them or not. But if not, they don't get something else. I'm not a short order cook.”

<p>Another unpopular advice is to wait until your child is very hungry before feeding them. “Wait until they are really hungry. Serve courses. Veggies first when that’s done second course is the thing they like.”</p>

Wait until they are very hungry

Another unpopular advice is to wait until your child is very hungry before feeding them. “Wait until they are really hungry. Serve courses. Veggies first when that’s done second course is the thing they like.”

<p>Another mom said blending the veggies in other foods helps make her kids eat vegetables. “Use an immersion blender when making sauce or chili. I would add roasted eggplant and spinach to Spaghetti sauce.”</p> <p>She also uses onions, garlic, carrots, tomatoes, jalapeño peppers, diced green chiles when making chili. “I blend the mixture with cooked meat (whichever you prefer: ground beef, ground Turkey, ground pork, ground chicken) until the mixture is somewhat smooth,” she shared. “Then I add beans. I added spinach to everything from meatballs, soup, and meatloaf.</p> <p>The same user aslo said she would make chowders with a variety of vegetable such as potatoes, corn, carrots, celery, onions, garlic, leeks, and bok choy.</p> <p>Are there other tricks you use to make your child eat vegetables besides the ones shared in <a href="https://www.reddit.com/r/AskReddit/comments/10zt2xi/parents_how_do_you_make_your_kids_eat_vegetables/?sort=top">this Reddit thread</a>? Share them in the comments!</p> <p><strong>Read Next:</strong></p> <p><span><a class="in-cell-link" href="https://arnienicola.com/family-dilemma-pregnant-woman-wants-sister-to-be-present-when-she-gives-birth-sister-wants-to-travel/" rel="noopener">Family Dilemma: Pregnant Woman Wants Sister To Be Present When She Gives Birth. Sister Wants To Travel</a><br> <a class="in-cell-link" href="https://arnienicola.com/woman-refuses-to-visit-her-in-laws-every-night-her-husband-gets-offended/" rel="noopener">Woman Refuses To Visit Her In-Laws Every Night. Her Husband Gets Offended</a><br> <a class="in-cell-link" href="https://arnienicola.com/mother-in-law-suggests-a-name-for-her-future-granddaughter-her-daughter-in-law-shuts-her-down/" rel="noopener">Mother-In-Law Suggests A Name For Her Future Granddaughter, Her Daughter-In-Law Shuts Her Down</a></span></p>

Blend them in other foods

Another mom said blending the veggies in other foods helps make her kids eat vegetables. “Use an immersion blender when making sauce or chili. I would add roasted eggplant and spinach to Spaghetti sauce.”

She also uses onions, garlic, carrots, tomatoes, jalapeño peppers, diced green chiles when making chili. “I blend the mixture with cooked meat (whichever you prefer: ground beef, ground Turkey, ground pork, ground chicken) until the mixture is somewhat smooth,” she shared. “Then I add beans. I added spinach to everything from meatballs, soup, and meatloaf.

The same user aslo said she would make chowders with a variety of vegetable such as potatoes, corn, carrots, celery, onions, garlic, leeks, and bok choy.

Are there other tricks you use to make your child eat vegetables besides the ones shared in this Reddit thread ? Share them in the comments!

how to solve how many combinations problems

She Took Her Daughter’s Makeup Away After Finding Out What She Did

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how to solve how many combinations problems

Stepfather Demands To Split Stepdaughter’s Inheritance From Deceased Father Equally Amongst Children

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how to solve how many combinations problems

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IMAGES

  1. Combinations

    how to solve how many combinations problems

  2. Combinations (video lessons, examples and solutions)

    how to solve how many combinations problems

  3. The Math Blog: Combinations (simple)

    how to solve how many combinations problems

  4. How to Solve Combinations Word Problems Quickly| Combinations and

    how to solve how many combinations problems

  5. Combination in Mathematics

    how to solve how many combinations problems

  6. How to Calculate Combination

    how to solve how many combinations problems

VIDEO

  1. Numerical On Laws of Chemical Combinations

  2. JNTUK MFCS R20 Previous Year Combinations Problems || Combinations Very Important Problems ||

  3. COMBINATION FORMULA- All subtraction combination formula # sums solving

  4. Combinations Problems #Combinations

  5. 11-MATHS |COMBINATIONS #3| BY DR ASHOK SIR #maths tricks #jee/cbse/ncert/state boards#maths shorts

  6. Solving Problems Part 1-Number Problems(Permutations and combinations)

COMMENTS

  1. How to Calculate Combinations: 8 Steps (with Pictures)

    To calculate combinations, you just need to know the number of items you're choosing from, the number of items to choose, and whether or not repetition is allowed (in the most common form of this problem, repetition is not allowed). Method 1 Calculating Combinations Without Repetition Download Article 1

  2. Combinations Calculator (nCr)

    The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Basically, it shows how many different possible subsets can be made from the larger set. For this calculator, the order of the items chosen in the subset does not matter. Factorial

  3. Counting, permutations, and combinations

    Math Statistics and probability Unit 8: Counting, permutations, and combinations 500 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test About this unit How many outfits can you make from the shirts, pants, and socks in your closet?

  4. Combinations (video lessons, examples and solutions)

    Solution: There are 6 players to be taken 2 at a time. Using the formula: They will need to play 15 games. Example: In a lottery, each ticket has 5 one-digit numbers 0-9 on it. a) You win if your ticket has the digits in any order. What are your changes of winning? b) You would win only if your ticket has the digits in the required order.

  5. Combination formula (video)

    Flag Miss H 9 years ago If the order doesn't matter, it is a combination. If the order does matter, it is a permutation. A permutation is an ordered combination. In this case, it doesn't matter what order the people are placed in to fill the chairs, it just matters which people you chose. 5 comments ( 317 votes) Upvote Downvote

  6. Combinations and Permutations

    1. Permutations with Repetition These are the easiest to calculate. When a thing has n different types ... we have n choices each time! For example: choosing 3 of those things, the permutations are: n × n × n (n multiplied 3 times) More generally: choosing r of something that has n different types, the permutations are: n × n × ... (r times)

  7. Combinations (practice)

    Do 4 problems. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

  8. Combination Calculator (nCr)

    Combination generator. This combination calculator (n choose k calculator) is a tool that helps you not only determine the number of combinations in a set (often denoted as nCr), but it also shows you every single possible combination (or permutation) of your set, up to the length of 20 elements. However, be careful!

  9. Combination Calculator (nCr Calculator)

    There are two formulas for calculating the number of possible combinations in an "n choose k" a.k.a. "n choose r" scenario, depending on whether repetition of the chosen elements is allowed or not. In both equations "!" denotes the factorial operation: multiplying the sequence of integers from 1 up to that number.

  10. Combinations

    Combinations. A combination is a way of choosing elements from a set in which order does not matter. A wide variety of counting problems can be cast in terms of the simple concept of combinations, therefore, this topic serves as a building block in solving a wide range of problems.

  11. Permutations and Combinations Problems

    Permutations and combinations are used to solve problems . Factorial Example 1: How many 3-digit numbers can you make using the digits 1, 2 and 3 without repetitions? method (1) listing all possible numbers using a tree diagram. We can make 6 numbers using 3 digits and without repetitions of the digits. method (2) counting:

  12. Combinations and Permutations Calculator

    Combinations and Permutations Calculator Find out how many different ways to choose items. For an in-depth explanation of the formulas please visit Combinations and Permutations. For an in-depth explanation please visit Combinations and Permutations. Power Users! You can now add "Rules" that will reduce the List:

  13. Using Combinations to Calculate Probabilities

    To calculate the number of combinations with repetitions, use the following equation. Where: n = the number of options. r = the size of each combination. The exclamation mark (!) represents a factorial. In general, n! equals the product of all numbers up to n. For example, 3! = 3 * 2 * 1 = 6. The exception is 0! = 1, which simplifies equations.

  14. Combination Calculator and Permutations Calculator

    This is 5 * 4 * 3 which can be written as 5!/2! (which is n! / (n - r)! with n=5, r=3). There is also an alternative way to pick a selection of 3 balls. Let's say we wanted to pick balls 123. Then we could go on to pick the remaining 2 balls too. This would give us the possible permutations 12345 and 12354.

  15. 12.2: Permutations and Combinations

    Consider the following counting problems: In how many ways can three runners finish a race? In how many ways can a group of three people be chosen to work on a project? What is the difference between these two problems?

  16. Easy Permutations and Combinations

    Combination: Choosing 3 desserts from a menu of 10. C (10,3) = 120. Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P (10,3) = 720. Don't memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are.

  17. Combinations Formula With Solved Example Questions

    Solution: Given, r = 4 (item sub-set) n = 18 (larger item) Therefore, simply: find "18 Choose 4" We know that, Combination = C (n, r) = n!/r! (n-r)! \ (\begin {array} {l}\frac {18!} {4! (18-4)!}=\frac {18!} {14!\times4!}\end {array} \) = 3,060 possible answers.

  18. Permutations & combinations (practice)

    Permutations & combinations. Google Classroom. You need to put your reindeer, Prancer, Quentin, Rudy, and Jebediah, in a single-file line to pull your sleigh. However, Rudy and Prancer are best friends, so you have to put them next to each other, or they won't fly.

  19. Combinations: Advanced Problems

    How to use the combination formula to solve problems that involve adding combinations of varying amounts. These combination problems are sometimes called 'less than' problems. ... Advanced Problems . To find out how many combinations of N objects taken either A or B at time, add both of the individual combinations N C a + N C b. Applied Example.

  20. Combinations and Permutations

    Solution: One way to solve this problem is to list all of the possible selections of 2 letters from the set of X, Y, and Z. They are: XY, XZ, and YZ. Thus, there are 3 possible combinations. Another approach is to use Rule 1. Rule 1 tells us that the number of combinations is n! / r!(n - r)!. We have 3 distinct objects so n = 3.

  21. Can you solve this viral math puzzle in 10 seconds? Test your IQ ...

    Quick IQ test: Here's how you solve it correctly Let's analyze the equation 2+5=12. Notice that if we add the result of the first sum to the numbers on the left, we arrive at the correct result.

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    5. Problem-solving and critical thinking. Being a good problem solver usually means knowing how to identify a problem and going through a series of steps to develop a solution. From entry-level employees up to your executives, those who can solve problems independently often become more critical thinkers, leading to better overall job performance.

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    Realm Cards in Nightingale represent all the aspects and elements that you can modify in a Realm. Realm Cards are an essential aspect of a Realmwalker's arsenal, as they make things easier for you ...

  24. Example: Combinatorics and probability (video)

    12 years ago While calculating # of hands with 1's , we use 32x31x30x29x28 as the numerator which seems obvious. However since the arrangement of 1's and the remaining cards does not matter , should the denominator not be 9 factorial ? • 3 comments ( 78 votes) Cameron Christensen 12 years ago No. I was initially confused here as well.

  25. Combinations Questions (With Solutions)

    The formula for Combination: To find the number of ways to select r items from n items without repetition, we have the formula for combination as follows: \ (\begin {array} {l}^ {n}C_ {r}=\binom {n} {r}=\frac {n!} {r! (n-r)!}\end {array} \) Learn more about Permutation and Combination. Video Lesson on Formulas for Combination 17,017

  26. Readers & Leaders: This is what's missing from your approach to problem

    In this edition of Readers & Leaders, sharpen your business problem-solving skills and learn ways to overcome friction, strengthen teams, and enhance project management efforts. After studying more than 2,000 teams, Robert I. Sutton shares friction-fixing tips to streamline processes for greater efficiency and less frustration. Andrew McAfee ...

  27. Solve Your Picky Eater Problems With These Tips On How To Make Your

    One user shared, "Don't be like my mom and boil down every vegetable until the consistency of mush. Boiled down, unsalted brussel sprouts tasted like crap.

  28. Probability using combinations (video)

    11 years ago I believe a factorial of a number is its product with all positive integers below. For example: 5! = 5 x 4 x 3 x 2 x 1 ( 34 votes)