how to solve permutation problems

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Precalculus

Course: precalculus   >   unit 8.

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Permutation formula

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How to Calculate Permutations

Last Updated: March 17, 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 71,792 times. Learn more...

If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. A permutation is an arrangement of objects in which the order is important [1] X Research source (unlike combinations , which are groups of items where order doesn't matter [2] X Research source ). You can use a simple mathematical formula to find the number of different possible ways to order the items. To start off, you just need to know whether repetition is allowed in your problem or not, and then pick your method and formula accordingly.

Calculating Permutations without Repetition

• For instance, you might be selecting 3 representatives for student government for 3 different positions from a set of 10 students. No student can be used in more than one position (no repetition), but the order still matters, since the student government positions are not interchangeable (a permutation where the first student is President is different from a permutation where they're Vice President).

• If you have a calculator handy, find the factorial setting and use that to calculate the number of permutations. If you're using Google Calculator, click on the x! button each time after entering the necessary digits.
• If you have to solve by hand, remember 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 example, you would calculate 10! by doing (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800 as a result. 7! would be (7 * 6 * 5 * 4 * 3 * 2 * 1), which would equal 5,040. You'd then calculate 3,628,800/5,040.
• In the example, you should get 720. That number means that, if you're picking from 10 different students for 3 student government positions, where order matters and there is no repetition, there are 720 possibilities.

Calculating Permutations with Repetition

• For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition.
• A permutation with repetition of n chosen elements is also known as an " n -tuple". [4] X Research source

Community Q&A

You Might Also Like

• ↑ https://www.mathsisfun.com/combinatorics/combinations-permutations.html
• ↑ https://betterexplained.com/articles/easy-permutations-and-combinations/
• ↑ https://www.learneroo.com/modules/10/nodes/62
• ↑ https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
• ↑ https://www.vitutor.com/statistics/combinatorics/permutations_repetition.html
• ↑ https://www.learneroo.com/modules/10/nodes/55
• https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

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Permutations

In these lessons, we will learn about permutations how to solve permutation with and without repetitions.

Related Pages Permutations Permutations and Combinations Permutations P(n,r) Permutations & Probability Probability

What Is Permutation?

In statistics, in order to find the number of possible arrangements of a set of objects, we use a concept called permutations. There are methods for calculating permutations, and it’s important to understand the difference between a set with and without repetition. Other important concepts that can apply to situations like permutations are the fundamental counting principal and basic probability.

How to find the number of permutations of a set?

Evaluating Factorials Use permutations to solve problems. Determine the number of permutations with indistinguishable items.

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

Counting Using Permutations Basic info on permutations and word problems using permutations are shown.

Permutations Involving Repeated Symbols - Example 1 This video shows how to calculate the number of linear arrangements of the word MISSISSIPPI (letters of the same type are indistinguishable). It gives the general formula and then grind out the exact answer for this problem.

Permutations Involving Repeated Symbols - Example 2 Basic info on permutations and word problems using permutations are shown.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Combinations and Permutations

What's the difference.

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.

"The combination to the safe is 472" . Now we do care about the order. "724" won't work, nor will "247". It has to be exactly 4-7-2 .

So, in Mathematics we use more precise language:

• When the order doesn't matter, it is a Combination .
• When the order does matter it is a Permutation .

In other words:

A Permutation is an ordered Combination.

Permutations

There are basically two types of permutation:

• Repetition is Allowed : such as the lock above. It could be "333".
• No Repetition : for example the first three people in a running race. You can't be first and second.

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)

(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)

Which is easier to write down using an exponent of r :

n × n × ... (r times) = n r

Example: in the lock above, there are 10 numbers to choose from (0,1,2,3,4,5,6,7,8,9) and we choose 3 of them:

10 × 10 × ... (3 times) = 10 3 = 1,000 permutations

So, the formula is simply:

2. Permutations without Repetition

In this case, we have to reduce the number of available choices each time.

Example: what order could 16 pool balls be in?

After choosing, say, number "14" we can't choose it again.

So, our first choice has 16 possibilites, and our next choice has 15 possibilities, then 14, 13, 12, 11, ... etc. And the total permutations are:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe we don't want to choose them all, just 3 of them, and that is then:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be arranged out of 16 balls.

Without repetition our choices get reduced each time.

But how do we write that mathematically? Answer: we use the " factorial function "

So, when we want to select all of the billiard balls the permutations are:

16! = 20,922,789,888,000

But when we want to select just 3 we don't want to multiply after 14. How do we do that? There is a neat trick: we divide by 13!

16 × 15 × 14 × 13 × 12 × ... 13 × 12 × ...   =  16 × 15 × 14

That was neat: the 13 × 12 × ... etc gets "cancelled out", leaving only 16 × 15 × 14 .

The formula is written:

Example Our "order of 3 out of 16 pool balls example" is:

(which is just the same as: 16 × 15 × 14 = 3,360 )

Example: How many ways can first and second place be awarded to 10 people?

(which is just the same as: 10 × 9 = 90 )

Instead of writing the whole formula, people use different notations such as these:

• P(10,2) = 90
• 10 P 2 = 90

Combinations

There are also two types of combinations (remember the order does not matter now):

• Repetition is Allowed : such as coins in your pocket (5,5,5,10,10)
• No Repetition : such as lottery numbers (2,14,15,27,30,33)

1. Combinations with Repetition

Actually, these are the hardest to explain, so we will come back to this later.

2. Combinations without Repetition

This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win!

The easiest way to explain it is to:

• assume that the order does matter (ie permutations),
• then alter it so the order does not matter.

Going back to our pool ball example, let's say we just want to know which 3 pool balls are chosen, not the order.

We already know that 3 out of 16 gave us 3,360 permutations.

But many of those are the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 are chosen. These are the possibilites:

So, the permutations have 6 times as many possibilites.

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

3! = 3 × 2 × 1 = 6

(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)

So we adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in their order any more):

That formula is so important it is often just written in big parentheses like this:

It is often called "n choose r" (such as "16 choose 3")

And is also known as the Binomial Coefficient .

All these notations mean "n choose r":

Just remember the formula:

n! r!(n − r)!

Example: Pool Balls (without order)

So, our pool ball example (now without order) is:

16! 3!(16−3)!

= 16! 3! × 13!

= 20,922,789,888,000 6 × 6,227,020,800

Notice the formula 16! 3! × 13! gives the same answer as 16! 13! × 3!

So choosing 3 balls out of 16, or choosing 13 balls out of 16, have the same number of combinations:

16! 3!(16−3)! = 16! 13!(16−13)! = 16! 3! × 13! = 560

In fact the formula is nice and symmetrical :

Also, knowing that 16!/13! reduces to 16×15×14, we can save lots of calculation by doing it this way:

16×15×14 3×2×1

Pascal's Triangle

We can also use Pascal's Triangle to find the values. Go down to row "n" (the top row is 0), and then along "r" places and the value there is our answer. Here is an extract showing row 16:

OK, now we can tackle this one ...

Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla .

We can have three scoops. How many variations will there be?

Let's use letters for the flavors: {b, c, l, s, v}. Example selections include

• {c, c, c} (3 scoops of chocolate)
• {b, l, v} (one each of banana, lemon and vanilla)
• {b, v, v} (one of banana, two of vanilla)

(And just to be clear: There are n=5 things to choose from, we choose r=3 of them, order does not matter, and we can repeat!)

Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out.

Think about the ice cream being in boxes, we could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and we will have 3 scoops of chocolate!

So it is like we are ordering a robot to get our ice cream, but it doesn't change anything, we still get what we want.

In fact the three examples above can be written like this:

So instead of worrying about different flavors, we have a simpler question: "how many different ways can we arrange arrows and circles?"

Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container).

So (being general here) there are r + (n−1) positions, and we want to choose r of them to have circles.

This is like saying "we have r + (n−1) pool balls and want to choose r of them". In other words it is now like the pool balls question, but with slightly changed numbers. And we can write it like this:

Interestingly, we can look at the arrows instead of the circles, and say "we have r + (n−1) positions and want to choose (n−1) of them to have arrows", and the answer is the same:

So, what about our example, what is the answer?

There are 35 ways of having 3 scoops from five flavors of icecream.

In Conclusion

Phew, that was a lot to absorb, so maybe you could read it again to be sure!

But knowing how these formulas work is only half the battle. Figuring out how to interpret a real world situation can be quite hard.

But at least you now know the 4 variations of "Order does/does not matter" and "Repeats are/are not allowed":

• Maths Questions

Permutation Questions

Permutation questions deal with the arrangement of objects in a specific order or formation of a number of different words from the letters of a given word, etc. There exist a variety of cases in which we apply permutations to get the possible results. In this article, you will get solved questions on permutation, and some practice questions for the same.

What is Permutation?

Permutation refers to the arrangement of objects in a definite order. That means permutation is the arrangement of objects in which order matters. The arrangement of r objects out of n objects can be calculated using the permutation formula. That is:

n P r = n!/(n – r)!

Learn in detail about permutation here.

Permutation Questions and Answers

1. Calculate the following:

(i) n P r when n = 12, r = 5

n P r = 12 P 5 = 12!/(12 – 5)!

= (12 × 11 × 10 × 9 × 8 × 7!)/7!

= 12 × 11 × 10 × 9 × 8

9 P 4 = 9!/(9 – 4)! = 9!/5!

= (9 × 8 × 7 × 6 × 5!)/5!

2. In how many different ways can the letters of the word THOUGHTS be arranged so that the vowels always come together?

Given word: THOUGHTS

Number of letters = 8

T’s = 2

H’s = 2

Number of vowels = 2 (O, U)

Vowels should come together.

So, the number of letters for arrangement = 7

i.e., (OU)THGHTS

Number of arrangements = 7!

And two vowels can be arranged in 2! ways.

Therefore, the total number of ways of arrangements = (7! × 2!)/(2! 2!)

= (7 × 6 × 5 × 4 × 3 × 2!)/2!

3. In how many ways can seven books be arranged on a shelf?

Number of ways in which the first book can be placed = 7

Number of ways in which the second book can be placed = 6

The total number of ways in which seven books can be arranged on a shelf = 7 × 6 × 5 × 4 × 3 × 2 × 1 (i.e., 7!)

4. How many different arrangements of letters of the word MATHEMATICS are possible?

Given word: MATHEMATICS

Number of letters = 11

Number of different arrangements = 11!/(2! 2! 2!)

= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1 × 2 × 1 × 2 × 1)

5. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?

Given digits: 1, 2, 3, 4, 6, 7

Number of digits = 6

Number of possible digits at unit’s place = 3 (2, 4 and 6)

⇒ Number of permutations = 3 P 1 = 3

When one of the digits is taken in units’ place, then the number of possible digits available = 5

⇒ Number of permutations = 5 P 2 = 5!/(5 – 2)! = 5!/3! = 120/6 = 20

The total number of permutations = 3 × 20 = 60.

Therefore, 60 three-digit numbers can be made using the given digits.

6. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Given: 5 men and 4 women

Total number of people = 9

The women occupy even places, which means they will be sitting in the 2nd, 4th, 6th and 8th places, whereas the men will be sitting in the 1st, 3rd, 5th, 7th and 9th places.

The number of arrangements in which 4 women can sit in 4 places = 4 P 4 = 4!/(4 – 4)! = 4!/0! = 24/1 = 24

5 men can occupy 5 seats in 5 ways.

That means the number of ways they can be seated = 5 P 5 = 5!/(5 – 5)! = 5!/0! = 120/1 = 120

Therefore, the total numbers of possible sitting arrangements = 24 × 120 = 2880

7. How many numbers are there between 100 and 1000 such that at least one of their digits is 7?

Total number of 3-digit numbers having at least one of their digits as 7 = (Total number of 3-digit numbers) – (Total number of 3-digit numbers in which 7 does not appear at all)

Let us find the total number of 3-digit numbers between 100 and 1000.

That means repetition of digits is allowed.

The hundred’s place can be filled in 9 ways, i.e., using digits from 1 to 9.

The ten’s and the unit’s place can be filled in 10 ways using the digits from 0 to 9.

∴ Total number of 3-digit numbers = 9 × 10 × 10 = 900

Now, we need to find the total number of 3-digit numbers in which 7 does not appear.

Here, 9 digits to be used, i.e., 0, 1, 2, 3, 4, 5, 6, 8, 9.

Now, the hundred’s place can be filled in 8 ways (excluding 0), and the tens’ and ones’ place can be filled in 9 ways each.

∴ Total number of 3-digit numbers in which 7 does not appear = 8 × 9 × 9 = 648

Hence, the numbers between 100 and 1000 in which at least one of their digits is 7 = 900 – 648 = 252

8. P, Q, R, S, and T sit on five chairs facing north. R will sit only on the leftmost chair, and Q will not sit anywhere to the left of P. In how many ways can they be seated?

R will sit on 1 and Q will sit somewhere to the right of P

R will sit on 1

Then there are three possible ways.

P on 2, so Q can be seated on 3, 4 or 5

The remaining two can be seated on two chairs in 2 ways

Number of possible ways = 3 × 2 = 6

P on 3, so Q can be seated on 4 or 5

Number of possible ways = 2 × 2 = 4

P on 4, so Q will be on 5

Number of possible ways = 2

Total number of possible ways = (6 + 4 + 2) = 12

9. In how many ways can 10 differently coloured beads be threaded on a string?

As the necklace can be turned over, clockwise and anti-clockwise arrangements are the same.

Also, the number of string arrangements of n objects = (n – 1)!/2

Therefore, the number of ways in which 10 differently coloured beads can be threaded on a string = (10 – 1)!/2 = 9!/2

10. How many 3 digit numbers can be formed with the digits 5, 6, 2, 3, 7 and 9, which are divisible by 5, and none of its digits are repeated?

If the number has unit’s digits as 0 or 5, then it will be divisible by 5.

Given digits: 5, 6, 2, 3, 7, 9

The digit that can be placed at the unit’s place so that the three-digit number is divisible by 5 is 5.

This can be done in 1 way.

Now, the tens and hundreds of places can be filled in 5 and 4 ways since the repetition of the digits are not allowed.

Therefore, the total number of such three digits numbers = 5 × 4 × 1 = 20

Practice Problems on Permutation

• How many ways can the letters of the word “EXAMINATION” be arranged such that the first and last letters are the same, and the vowels are together?
• How many numbers between 100 and 1000 use only odd digits, no digit being repeated?
• How many 3-letter words, with or without meaning, can be formed out of the letters of the word LOGARITHMS, if repetition of letters is not allowed?
• In how many ways can 9 different colour balls be arranged in a row so that the black, white, red and green balls are never together?
• Find the sum of all the possible numbers of 4 digits formed by digits 3, 5, 5, and 6 using each digit once.

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7.3.1: Permutations (Exercises)

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• Rupinder Sekhon and Roberta Bloom
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Do the following problems using permutations.

Free Mathematics Tutorials

Permutations and Combinations Problems

Permutations and combinations are used to solve problems .

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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How to Solve Permutations and Combinations? (+FREE Worksheet!)

Learn how to solve mathematics word problems containing Permutations and Combinations using formulas.

Related Topics

• How to Interpret Histogram
• How to Interpret Pie Graphs
• How to Solve Probability Problems
• How to Find Mean, Median, Mode, and Range of the Given Data

Step by step guide to solve Permutations and Combinations

• Permutations: The number of ways to choose a sample of \(k\) elements from a set of \(n\) distinct objects where order does matter, and replacements are not allowed. For a permutation problem, use this formula: \(\color{blue}{_{n}P_{k }= \frac{n!}{(n-k)!}}\)
• Combination: The number of ways to choose a sample of \(r\) elements from a set of \(n\) distinct objects where order does not matter, and replacements are not allowed. For a combination problem, use this formula: \(\color{blue}{_{ n}C_{r }= \frac{n!}{r! (n-r)!}}\)
• Factorials are products, indicated by an exclamation mark. For example, \(4!\) Equals: \(4×3×2×1\). Remember that \(0!\) is defined to be equal to \(1\).

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Permutations and combinations – example 1:.

How many ways can the first and second place be awarded to \(10\) people?

Since the order matters, we need to use the permutation formula where \(n\) is \(10\) and \(k\) is \(2\). Then: \(\frac{n!}{(n-k)!}=\frac{10!}{(10-2)!}=\frac{10!}{8!}=\frac{10×9×8!}{8!}\), remove \(8!\) from both sides of the fraction. Then: \(\frac{10×9×8!}{8!}=10×9=90\)

Permutations and Combinations – Example 2:

How many ways can we pick a team of \(3\) people from a group of \(8\)?

Since the order doesn’t matter, we need to use a combination formula where \(n\) is \(8\) and \(r\) is \(3\). Then: \(\frac{n!}{r! (n-r)!}=\frac{8!}{3! (8-3)!}=\frac{8!}{3! (5)!}=\frac{8×7×6×5!}{3! (5)!}\), remove \(5!\) from both sides of the fraction. Then: \(\frac{8×7×6}{3×2×1}=\frac{336}{6}=56\)

Exercises for Solving Permutations and Combinations

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Calculate the value of each..

• \(\color{blue}{4!=}\)
• \(\color{blue}{4!×3!=}\)
• \(\color{blue}{5!=}\)
• \(\color{blue}{6!+3!=}\)
• There are \(7\) horses in a race. In how many different orders can the horses finish?
• In how many ways can \(6\) people be arranged in a row?

Download Combinations and Permutations Worksheet

• \(\color{blue}{24}\)
• \(\color{blue}{144}\)
• \(\color{blue}{120}\)
• \(\color{blue}{726}\)
• \(\color{blue}{5,040}\)
• \(\color{blue}{720}\)

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by: Effortless Math Team about 4 years ago (category: Articles , Free Math Worksheets )

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Permutations With Restriction

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A permutation is an ordering of a set of objects. When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions . Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Other common types of restrictions include restricting the type of objects that can be adjacent to one another, or changing the ordering mechanism from a line to another topology (e.g. a round table instead of a line, or a keychain instead of a ring).

Problems of this form are perhaps the most common in practice. For example, deciding on an order of what to eat, do, or watch are all implicit examples of permutations with restrictions, since it is obviously impractical to plan an ordering for all possible foods/tasks/shows.

Restriction to Few Objects

Restrictions on adjacent objects, restrictions on topology.

Restrictions to few objects is equivalent to the following problem:

Given \(n\) distinct objects, how many ways are there to place \(k\) of them into an ordering?

As in the strategy for dealing with permutations of the entire set of objects, consider an empty ordering which consists of \( k\) empty positions in a line to be filled by \(k\) objects. There are \( n\) choices for which of the \(n\) objects to place in the first position. After the first object is placed, there are \(n-1\) remaining objects, so there are \( n-1\) choices for which object to place in the second position. Repeating this argument, there are \( n-2\) choices for the third position, \( n-3\) choices for the fourth position, and so on. Finally, for the \( k^\text{th}\) position, there are \( n - (k-1) = n- k + 1\) choices. Then the rule of product implies the total number of orderings is given by the following:

Given \( n \) distinct objects, the number of different ways to place \(k\) of them into an ordering is \[ P^n_k = n (n-1)(n-2) \cdots (n-k+1) = \frac{n!}{(n-k)!} . \] This is also known as a \(k\)-permutation of \(n\).

Lisa has 12 ornaments and wants to put 5 ornaments on her mantle. How many ways can she do this? By the rule of product , Lisa has 12 choices for which ornament to put in the first position, 11 for the second, 10 for the third, 9 for the fourth and 8 for the fifth. So the total number of choices she has is \( 12 \times 11 \times 10 \times 9 \times 8 \). Using the factorial notation, the total number of choices is \( \frac{12!}{7!} \). \(_\square\)

Out of a class of 30 students, how many ways are there to choose a class president, a secretary, and a treasurer? A student may hold at most one post. Solution 1: We can choose from among 30 students for the class president, 29 students for the secretary, and 28 students for the treasurer. Hence, by the rule of product, the number of possibilities is \( 30 \times 29 \times 28 = 24360 \). Solution 2: By the above discussion, there are \( P_{27}^{30} = \frac {30!}{(30-3)!} \) ways. While it is extremely hard to evaluate \( 30!\) and \( 27!\), we notice that dividing out gives \( 30 \times 29 \times 28 = 24360 \). \(_\square\)

A team of explorers are going to randomly pick 4 people out of 10 to go into a maze. How many different ways are there to pick?

Then the 4 chosen ones are going to be separated into 4 different corners: North, South, East, West. How many ways can they be separated?

How many different ways are there to color a \(3\times3\) grid with green, red, and blue paints, using each color 3 times?

Try other painting \(n\times n\) grid problems.

Lisa has 4 different dog ornaments and 6 different cat ornaments that she wants to place on her mantle. All of the dog ornaments should be consecutive and the cat ornaments should also be consecutive. How many ways can they be arranged? We have to decide if we want to place the dog ornaments first, or the cat ornaments first, which gives us 2 possibilities. We can arrange the dog ornaments in \( 4!\) ways, and the cat ornaments in \( 6!\) ways. Hence, by the rule of product , there are \( 2 \times 6! \times 4! =34560 \) ways to arrange the ornaments. \(_\square\)

How many arrangements are there of the letters of BANANA such that no two N's appear in adjacent positions?

9 different books are to be arranged on a bookshelf. 4 of these books were written by Shakespeare, 2 by Dickens, and 3 by Conrad. How many possible permutations are there if the books by Conrad must be separated from one another?

6 friends go out for dinner. How many ways are there to sit them around a round table? Rotations of a sitting arrangement are considered the same, but a reflection will be considered different. Solution 1: Since rotations are considered the same, we may fix the position of one of the friends, and then proceed to arrange the 5 remaining friends clockwise around him. Thus, there are \( 5! = 120 \) ways to arrange the friends. Solution 2: There are \( 6! \) ways to seat the 6 friends around the table. However, since rotations are considered the same, there are 6 arrangements which would be the same. Hence, to account for these repeated arrangements, we divide out by the number of repetitions to obtain that the total number of arrangements is \( \frac {6!}{6} = 120 \). Both solutions are equally valid and illustrate how thinking of the problem in a different manner can yield another way of calculating the answer. \(_\square\)

• Permutations
• Permutations with Repetition

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Can AI Solve the Math Mysteries Stumping the Field’s Brightest Minds?

Artificial Intelligence already won a gold ribbon in the most elite high school math competition. It could help humans conduct boundary-pushing math research next.

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Each year since 1959, high school math students from more than 100 countries have competed to solve a wide variety of math problems involving algebra , geometry , and number theory quickly and elegantly. Many IMO winners have secured prestigious math awards as adults, including the coveted Fields Medal .

In essence, IMO is a benchmark for students to see if they have what it takes to succeed in the field of mathematics. Now, artificial intelligence has aced the test—well, the geometry part at least.

In a paper published this January in Nature , a team of scientists from Google’s DeepMind have introduced a new AI called AlphaGeometry that’s capable of passing the geometry section of the International Math Olympiad without relying on human examples.

“We’ve made a lot of progress with models like ChatGPT … but when it comes to mathematical problems, these [large language models] essentially score zero,” Thang Luong , Ph.D., a senior staff research scientist at Google DeepMind and a senior author of the AlphaGeometry paper, tells Popular Mechanics . “When you ask [math] questions, the model will give you what looks like an answer, but [it actually] doesn’t make sense.”

For example, things get messy when AI tries to solve an algebraic word problem or a combinatorics problem that asks it to find the number of permutations (or versions) of a number sequence.

To answer math questions of this caliber , AlphaGeometry relies on a combination of symbolic AI—which Luong describes as being precise but slow—and a neural network more similar to large language models (LLMs) that is responsible for the quick, creative side of problem-solving.

Yet, math experts aren’t convinced that an AI made to solve high school-level math problems is ready to take off the training wheels and tackle more difficult subjects, e.g. advanced number theory or combinatorics, let alone boundary-pushing math research.

Why AI Struggles With Math

While LLM-powered AI tools have exploded in the past two years, these models have routinely struggled to handle math problems. This is part of what makes AlphaGeometry stand out from the crowd. But even so, that doesn’t necessarily mean it’s ready to tackle higher-level math yet.

.css-2l0eat{font-family:UnitedSans,UnitedSans-roboto,UnitedSans-local,Helvetica,Arial,Sans-serif;font-size:1.625rem;line-height:1.2;margin:0rem;padding:0.9rem 1rem 1rem;}@media(max-width: 48rem){.css-2l0eat{font-size:1.75rem;line-height:1;}}@media(min-width: 48rem){.css-2l0eat{font-size:1.875rem;line-height:1;}}@media(min-width: 64rem){.css-2l0eat{font-size:2.25rem;line-height:1;}}.css-2l0eat b,.css-2l0eat strong{font-family:inherit;font-weight:bold;}.css-2l0eat em,.css-2l0eat i{font-style:italic;font-family:inherit;} “The challenge of AI is that [it] cannot come up with new concepts.”

Marijin Heule , Ph.D., is an associate professor of computer science at Carnegie Mellon University whose work focuses on another kind of automated theorem prover called SAT solvers. In this case, “SAT” refers to a measure of validity called “satisfiability” and not the math section of the high school SAT.

“When it comes down to solving math problems or problems in general, the challenge of AI is that [it] cannot come up with new concepts,” Heule tells Popular Mechanics .

This limitation impacts symbolic AI and neural networks in different ways, Heule explains, but both stem from the issue that these AI rely on an existing bank of human knowledge. However, this isn’t exactly true for AlphaGeometry because it relies on synthetic data , which isn’t based on human examples but is made to mimic them.

While AIs might not be effective mathematicians on their own, that doesn’t necessarily mean they can’t be great apprentices to human mathematicians.

“At least for the foreseeable future, [AI will] be mostly assisting,” Heule says. “One of the other things that these machines can do really well is they can tell you if there is an incorrect argument and [offer] a counterexample.”

These AI-powered nudges can help researchers distinguish research dead-ends from promising paths.

Why Geometry?

Of all the math fields the AlphaGeometry team could have tackled, Luong says there were a few factors that helped them zero in on geometry .

“I think geometry is visually appealing [and] we do geometry as kids,” he says. “And geometry is everywhere in design and architecture, so it’s very important.”

Geometry also offered a unique challenge as being one of the International Math Olympiad fields with the fewest number of proof examples written in a computer-friendly format (e.g. without pictures).

While Heule agrees that AlphaGeometry is “really cool work,” he admits that designing a geometry solver is one of the easier tasks for a math AI.

While human computer scientists did work behind the scenes to formalize geometry problems in a way that computers can reason about, Heule says the reasoning is pretty straightforward once that preparation work is complete.

In part, this is because the considerations of geometry problems (e.g. the relationship between angles, points, and lines) are fairly contained compared to more complex areas, he says.

Take for example Fermat’s Last Theorem . This number theory problem took over three centuries to solve, and Heule says it would be extremely difficult to explain its solution to AI, let alone ask AI to solve it.

“Large-scale fields of modern mathematics … are so big that any one of them contains multitudes,” says Heather Macbeth , Ph.D., an assistant professor of mathematics at Fordham University with a focus on geometry. “I think, maybe a more precise question would be to talk about the styles of problems, which might occur within any mathematical field that some of these AI systems are useful for,” she tells Popular Mechanics.

For example, AI could be useful for pattern recognition or so-called needle-in-a-haystack problems where mathematicians are looking for something with a very particular property, Macbeth says.

Toward General AI

While AI likely won’t be solving centuries-old math problems in the near future, Luong is confident there are still existing advancements on the horizon for AlphaGeometry and its ilk. Perhaps these models could even graduate high school and take on the Putnam Mathematical Competition for undergraduate students.

But beyond math tests themselves, Luong is hopeful about what models like AlphaGeometry could mean for the field of AI at large—in particular, researchers’ goals of designing a generalized AI.

“If we want to talk about building an artificial general intelligence, where we want the AI to be as smart as a human, I think the AI needs to be able to perform deep reasoning,” Luong says. “This means that the AI needs to be able to plan ahead for many, many steps [and] see the big picture of how things connect together … the IMO is the perfect test for that.”

Sarah is a science and technology journalist based in Boston interested in how innovation and research intersect with our daily lives. She has written for a number of national publications and covers innovation news at Inverse .

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• Custom test

As soon as everyone in the camp fell asleep, Kirill sneaked out of the tent and went to the Wise Oak to gather mushrooms.

It is known that there are $$$n$$$ mushrooms growing under the Oak, each of which has magic power $$$v_i$$$. Kirill really wants to make a magical elixir of maximum strength from the mushrooms.

The strength of the elixir is equal to the product of the number of mushrooms in it and the minimum magic power among these mushrooms. To prepare the elixir, Kirill will sequentially pick one mushroom growing under the Oak. Kirill can gather mushrooms in any order.

However, it's not that simple. The Wise Oak informed Kirill of a permutation of numbers $$$p$$$ from $$$1$$$ to $$$n$$$. If Kirill picks only $$$k$$$ mushrooms, then the magic power of all mushrooms with indices $$$p_1, p_2, \dots, p_{k - 1}$$$ will become $$$0$$$. Kirill will not use mushrooms with zero magic power to prepare the elixir.

Your task is to help Kirill gather mushrooms in such a way that he can brew the elixir of maximum possible strength. However, Kirill is a little scared to stay near the oak for too long, so out of all the suitable options for gathering mushrooms, he asks you to find the one with the minimum number of mushrooms.

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ different integers from $$$1$$$ to $$$n$$$ in any order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears in the array twice) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$, but $$$4$$$ appears in the array).

Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 200\,000$$$) — the number of mushrooms.

The second line contains an array $$$v$$$ of size $$$n$$$ ($$$1\le v_i \le 10^9$$$) — the magic powers of the mushrooms.

The third line contains a permutation $$$p$$$ of numbers from $$$1$$$ to $$$n$$$.

It is guaranteed that the sum of the values of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

For each test case, output two integers separated by a space — the maximum strength of the elixir that can be brewed and the minimum number of mushrooms that Kirill needs to use for this.

In the first example, you need to take the mushrooms with indices $$$1$$$ and $$$2$$$, so the strength of the elixir is equal to $$$2 \cdot \min(a_1, a_2) = 2 \cdot \min(9, 8) = 2 \cdot 8 = 16$$$. Note that the magic power of the mushroom with index $$$3$$$ after picking two mushrooms will become $$$0$$$.