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Permutation refers to the various ways in which a set of elements can be ordered or arranged. On the other hand, combinations deal with the number of ways to choose a subset from a larger set, without considering the order of selection. Permutations concept is crucial in solving problems related to arranging items, such as seating arrangements, code combinations, or permutations of a sequence while combinations are particularly useful when you're interested in forming groups. At Smartkeeda, you can find important permutation and combination questions with answers . By practicing with these questions, you can strengthen your grasp on the concepts and improve your problem-solving skills.

Additionally, we offer a convenient permutation and combination questions PDF for free download, allowing you to access a variety of Permutation and combination Questions with Answers anytime, anywhere. Permutation and combination questions at Smartkeeda include memory-based questions from various exams. Each question is accompanied by a step-wise approach ensuring a clear understanding of the underlying principles. Additionally, we have included tricks and shortcuts wherever applicable, helping you optimize your problem-solving approach. By practicing permutation and combination questions with answers at Smartkeeda, you not only familiarize yourself with the types of questions that may appear in exams but also develop the skills to solve them in minimal time without compromising accuracy.

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  • Permutation and Combination Practice Questions

Permutation and combination is a very important topic in any competitive exams. We have covered this topic and all its sections in our earlier articles. Today we are going to discuss the permutation and combination practice questions. There are different types of practice questions for you to practice and get ready for the competitive exams.

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Permutation of ‘x’ things using all of them

Directions: For the questions in the section you need to find the distinctive ways to find the answer.

1. Using all the letters of the word GIFT how many distinct words can be formed?

A. 22 words            B. 24 words        C. 256 words       D. 200 words

2. Find out how many distinct three-digit numbers can be formed using all the digits of 1, 2, and 3.

A. 4            B. 5           C. 6              D. 7

3. In how many different ways can five friends sit for a photograph of five chairs in a row?

A. 120 ways       B. 24 ways         C. 240 ways            D. 720 ways

4. In how many different ways can the letters of the word MAGIC can be formed?

A. 24 ways           B. 120 ways           C. 240 ways              D. 720 ways

5. For the above word how many different types of arrangement are possible so that the vowels are always together?

A. 44 words       B. 24 words            C. 48 words           D. 60 words

6. In how many ways can the letters of the word BEAUTY be arranged?

A. 360           B. 5!           C. 6!             D. 7!

7. For the above word, if the vowels are always together than how many types of arrangement can be possible?

A. 4! * 3!       B. 6!        C. 4!           D. 4! * 3

8. A person has 4 coins if different denominations. What is the number of different sums of money the person can form?

A. 12        B. 15              C. 11             D. 16

permutation and combination practice questions

3. A  120 ways

4. B 120 ways

5. C 48 words

7. A 4! * 3!

Permutations of n things taking some of them at one time and when some things are alike

Directions: Answer the questions based on the data given to you

1. If repetition is not allowed then how many distinct three-digit numbers can be formed using the digits (1, 2, 3, 4, 5)?

A. 60 ways          B. 50 ways              C. 40 ways           D. 30 ways

2. Find out the distinct four-letter words that can be formed using the word SINGAPORE.

A. 256       B. 1024            C. 3024            D. 2048

3. Find out how many distinct three-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 such that the digits are in ascending order.

A. 80          B. 81              C. 83            D. 84

4. How many can 3 digits be formed using the digits from 1 to 5 if the digit 2 is never there in the number?

A. 24           B. 36         C. 40            D. 52

5. If no repetition is not allowed then how many numbers between 2000 and 3000 can be formed using the digits from 0 to 7?

A. 42             B. 336             C. 210              D. 440

6. In how many ways can Kamal choose a consonant and a vowel from the letters of the word ALLAHABAD?

A. 4             B. 5              C. 6           D. 9

7. Find out the number of distinctive words that can be formed using the word GOOD.

A. 16          B. 24             C. 28           D. 48

8. How many different words can be formed from the alphabets of the word SCISSORS?

A. 1440         B. 1680               C. 1800             D. 2100

9. How many distinct words can be formed using the word MINIMUM?

A. 420           B. 450            C. 1024            D. 1048

1. A 60 ways

Permutation when the repetition of the words are allowed

Directions: The questions in this section consists of the repetition of the words or letters or numbers or alphabets.

1. If repetition is allowed then how many different three digits numbers can be formed using the digits from 1 to 5?

A. 125         B. 27          C. 120           D. None of the above

2. In how many ways can two letters be selected from the English alphabet if repetition is allowed?

A. 650         B. 325           C. 52          D. 51

3. Priya has five friends in how many ways can she invited five or more friends for dinner?

A. 6         B. 7           C. 15            D. 21

4. When repetition is allowed, how many numbers between 2000 and 4000 can be selected from the digits 1 to 5?

A. 248          B. 249          C. 250               D. 128

5. How many can four digits be formed using the digits 0, 1, 2, and 3 (repetition is allowed)?

A. 12          B. 24           C. 256          D. 192

6. In a word jumble, there are 8 consonants and 5 vowels given. Find out in how many ways can we form a 5-letter word having three consonants and 2 vowels?

A. 67200            B. 8540           C. 720           D. None of these

7. There are 45 games in total in a competition. Many teams took part in the competition and each of them must play one with the other teams. In total how many teams took part in the competition?

A. 5           B. 10            C. 15            D. 20

Different Types of Arrangement

Directions: In this section, you need to find out how many different types of arrangements are possible.

1. In how many ways can you select a diamond or a king from a pack of cards?

A. 16             B. 20          C. 24            D. 8

2. A circular table has 6 chairs, out of this 6, five are identical. In how many ways can the six people be arranged on these chairs?

A. 120          B. 720            C. 360              D. 60

3. Jay invited 10 of his friends on his birthday. If all of them greeted each other with a handshake then how many handshakes will take place?

A. 90            B. 110             C. 45           D. 55

4. In how many can the 4 couples sit around a circular table so that no two men are sitting together?

A. 7!          B. 6!             C. 3! * 4!            D. 3! * 3!

5. There are three dice each of them having faces with a number from 1 to 6. These dices are rolled. Find the number of possible outcomes such that at least one of the dice shows the number 2.

A. 36         B. 91             C. 81             D. 116

4. C 3! * 4!

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1000+ Permutation and Combination Question Bank Pdf - 1

Model MCQ Online Test

Question: 1

In how many different ways can the letters of the word ‘GAMBLE’ be arranged?

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Question: 2

In how many different ways can the letters of the word ‘SMART’ be arranged?

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Question: 3

In how many different ways can the letters of the word ‘TOTAL’ be arranged?

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Question: 4

In how many different ways can the letters of the word ‘DAILY’ be arranged?

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Question: 5

In how many different ways can the letters of the word ‘AWARE’ be arranged?

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12.2: Permutations and Combinations

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  • Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
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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.

  • JEE Main Permutations And Combinations Previous Year Questions With Solutions

JEE Main Permutations and Combinations Previous Year Questions With Solutions

JEE Main Mathematics Permutations and Combinations previous year papers questions with solutions are available here. Find the permutation and combination questions from the previous years of IIT JEE Main in this article, along with a detailed explanation for each question. These questions will help you analyze the type and difficulty level of the questions in the JEE Main question papers and how to solve them with the right approach.

Download the list of questions as PDF:-

Download PDF of JEE Main Previous Year Solved Questions on Permutations and Combinations

Other important concepts in probability.

  • Binomial Theorem
  • Combinations
  • Permutation

JEE Main Maths Permutations and Combinations Previous Year Questions With Solutions

Question 1: A letter lock consists of three rings, each marked with ten different letters. In how many ways is it possible to make an unsuccessful attempt to open the lock?

Two rings may have the same letter at a time, but the same ring cannot have two letters at a time. Therefore, we must proceed ring wise. Each of the three rings can have any one of the 10 different letters in 10 ways.

Therefore, the total number of attempts = 10 × 10 × 10 = 1000.

But out of these 1000 attempts, only one attempt is successful.

Therefore, the required number of unsuccessful attempts = 1000 – 1 = 999.

Question 2: The total number of positive integral solution for x, y, z such that x* y * z = 24, is ________.

x* y * z = 24

x* y * z = 2 3 × 3 1

The number of ways of distributing ‘n’ identical balls into ‘r’ different boxes is (n + r − 1) C (r − 1)

Here we have to group 4 numbers into three groups

Number of integral positive solutions

= (3 + 3 − 1) C (3 − 1) × (1+ 3 − 1) C (3 − 1)

= 5 C 2 × 3 C 2

Question 3: Find the total number of signals that can be made by five flags of a different colour when any number of them may be used in any signal.

Case I: When only one flag is used. No. of signals made = 5 P 1 = 5.

Case II: When only two flags are used. Number of signals made = 5 P 2 = 5 * 4 = 20.

Case III: When only three flags are used. Number of signals is made = 5 P 3 = 5 * 4 * 3 = 60.

Case IV : When only four flags are used. Number of signals made = 5 P 4 = 5 * 4 * 3 * 2 = 120.

Case V : When five flags are used. Number of signals made = 5 P 5 = 5! = 120.

Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.

Question 4: Prove that if each of the ‘m’ points in one straight line is joined to each of the n points on the other straight line, excluding the points on the given two lines. The number of points of intersection of these lines is \(\begin{array}{l}\frac{1}{4}mn(m – 1)(n – 1).\end{array} \)

To get one point of intersection, we need two points on the first line and two points on the second line. These can be selected from n-points in n C 2 ways and m points in m C 2 ways.

Therefore, the required number = m C 2 × n C 2 = (m(m-1))/2!  × (n(n-1))/2!

Question 5: There are ten points in a plane. Of these ten points, four points are in a straight line, and except for these four points, no other three points are in the same straight line. Find

(i) The number of straight lines formed.

(ii) The number of triangles formed.

(iii) The number of quadrilaterals formed by joining these ten points.

(i) For a straight line, we need 2 points.

(In the last case only one straight line is formed)

Therefore, the required number = 15 + 24 + 1 = 40

(ii) For a triangle, we need 3 points.

(In the last case number of triangles formed is 0)

Therefore, the required number = 20 + 60 + 36 + 0 = 116

(iii) For a quadrilateral, we need 4 points.

Therefore, the required number = 15 + 80 + 90 = 185.

Question 6: Two numbers are chosen from 1, 3, 5, 7,… 147, 149 and 151 and multiplied together in all possible ways. The number of ways which will give the product a multiple of 5 is ______.

In the numbers 1, 3, 5, 7,…, 147, 149, 151, the numbers multiples of 5 are 5, 15, 25, 35, …, 145, which form an arithmetic sequence.

T n = a + (n − 1) × d

145 = 5 + (n – 1) × 10

and if the total number of terms in the given sequence is m, then

151 = 1 + (m – 1) × 2

So, the number of ways in which a product is a multiple of 5 = (both two numbers from 5, 10, 15, 20, 25, 30, 35, … , 150) or (one number from 5, 10, 15, 20, 25, 30, 35, …, 150 and one from remaining numbers)

= 15 C 2 + 15 C 1 × (76 – 15) C 1

= 105 + 15 × 61

= 105 + 915

Question 7: If the letters of the word SACHIN are arranged in all possible ways, and these words are written in a dictionary, at what serial number does the word SACHIN appear?

(1) Alphabetical order is A, C, H, I, N, S

No. of words starting with A – 5!

No. of words starting with C – 5!

No. of words starting with H – 5!

No. of words starting with I – 5!

No. of words starting with N – 5!

Hence, the required number = 5! + 5! + 5! + 5! + 5! + 1 = 120+ 120 + 120 + 120 + 120 + 1 = 601.

Question 8:   \(\begin{array}{l}\text{The value of}\ ^{50}C_4 + \sum_{r=1}^{6} {}^{56 – r} C_3\ \text{is}—–.\end{array} \)

= 50 C 4 + 55 C 3 + 54 C 3 + 53 C 3 + 52 C 3 + 51 C 3 + 50 C 3

= ( 50 C 4 + 50 C 3 ) + 55 C 3 + 54 C 3 + 53 C 3 + 52 C 3 + 51 C 3

= ( 51 C 4 + 51 C 3 ) + 55 C 3 + 54 C 3 + 53 C 3 + 52 C 3

= ( 52 C 4 + 52 C 3 ) + 55 C 3 + 54 C 3 + 53 C 3

= ( 53 C 4 + 53 C 3 ) + 55 C 3 + 54 C 3

= ( 54 C 4 + 54 C 3 ) + 55 C 3

= 55 C 4 + 55 C 3 

Question 9: Assuming the balls to be identical except for the difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is ______.

Number of white balls = p = 10

Number of green balls = q = 9

Number of black balls = r = 7

Total ways of selection = (p + 1) * (q + 1) * (r + 1) – 1

= [11 * 10 * 8] – 1

Question 10: From 6 different novels and 3 different dictionaries, 4 novels and a dictionary are to be selected and arranged in a row on the shelf such that the dictionary is in the middle. What is the number of such arrangements?

4 novels can be selected from 6 novels in 6 C 4 ways.

1 dictionary can be selected from 3 dictionaries in 3 C 1 ways.

As per the given, the dictionary selected is fixed in the middle.

Therefore, the remaining 4 novels can be arranged in 4! ways.

Hence, the required number of ways of arrangement = 6 C 4 * 3 C 1. * 4! = 1080

Question 11: Number of divisors of n = 38808 (except 1 and n) is _____.

Since, 38808 = 8 × 4851

= 8 × 9 × 539

= 8 × 9 × 7 × 7 × 11

= 2 3 × 3 2 × 7 2 × 11

So, the number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72.

This includes two divisors, 1 and 38808.

Hence, the number of divisors except 1 and n = 72 – 2 = 70.

Question 12: An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is _______.

Since any of the digits 2, 5 and 7 can be used at any place, the total number of such positive n-digit numbers are 3 n .

Since we have to form 900 distinct numbers, 3 n ≥ 900 ⇒ n = 7.

Question 13: In how many ways can 15 members of a council sit around a circular table when the Secretary is to sit on one side of the Chairman and the Deputy Secretary on the other side?

Since the total members are 15, one is to the left because of the circular condition.

Therefore, the remaining members are 14, but three special members constitute a member.

Therefore the required number of arrangements is 12! ×2 because the chairman remains between the two specified persons, and the person can sit in two ways.

Question 14: In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?

Since the total number of ways in which boys can occupy any place is (5 − 1)! = 4!

5 girls can sit accordingly in 5! ways.

Hence the required number of ways = 4! × 5 !

Question 15: A five-digit number divisible by 3 has to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is ________.

We know that a five-digit number is divisible by 3, if and only if the sum of its digits (= 15) is divisible by 3.

Therefore, we should not use 0 or 3 while forming the five-digit numbers.

(i) In this case, we do not use 0; the five-digit number can be formed (from the digit 1, 2, 3, 4, 5) in 5 P 5 ways.

(ii) In this case, we do not use 3; the five-digit number can be formed (from the digit 0, 1, 2, 4, 5) in (4 × 4 × 3 × 2 × 1) ways = 96 ways

The total number of such 5 digit number = 5 P 5 + 96

Permutation and Combination JEE Advanced Previous Year Questions With Solutions

How Do IIT JEE Permutations and Combinations Question Papers Help Students?

IIT JEE aspirants can find Permutations and Combinations JEE Main Maths previous year questions with solutions here. This set of questions helps aspirants to understand the type of questions asked in the JEE exams. Download JEE Main previous year Permutations and Combinations questions PDFs for free and crack JEE exam.

Permutations and Combinations MCQs

Permutations and Combinations MCQs

Permutations and Combination Problems and Solutions

Permutations and Combination Problems and Solutions

Problems based on Permutations and Combinations

Problems based on Permutations and Combinations

JEE Permutations and Combinations Solved Questions

JEE Permutations and Combinations Solved Questions

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

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  • permutation and combination tricks and shortcuts

Permutation and Combination Tricks and Shortcuts PDF For Bank PO, Clerk Exams

Nov 16 2023

permutation and combination question bank pdf

Permutation and Combination Tricks and Shortcuts For Bank Exams: Permutation and combination tricks pdf for bank exams is added in this permutation and combinations tricks for bank exams article. Candidates who are effectively preparing for their upcoming bank exams and other competitive exams like ssc chsl, ssc cgl, ssc mts, railways and more shall utilize this permutation and combination tricks for bank exams to score maximum marks in the government exams. Permutation and combination are some of the most interesting topics in the quantitative aptitude section of the bank exams. In most of the bank exams like sbi clerk, sbi po, ibps clerk, ibps po, rrb clerk, rrb po and other insurance exams permutation and combination questions play a major role. So aspirants must be proficient with the permutation and combination tricks for bank exams. This permutation and combination tricks for bank exams will help you to solve a variety of questions simply. The aspirants who want to maximise their scores in the bank exams can use this permutation and combination tricks to solve the questions as soon as possible. We have also attached permutation and combination tricks pdf in this permutation and combination tricks for bank exams article. Make use of these permutation and combination tricks pdf to get more marks in the bank exams.

Permutation and Combination Tricks and Shortcuts PDF For Bank Exams

We have uploaded the permutation and combination tricks and shortcuts pdf for bank exams in this permutation and combination tricks for bank exams article. Candidates can utilize these permutation and combination tricks to solve different types of questions simply with speed and accuracy. You can download the permutation and combination tricks and shortcuts pdf for your future reference. 

Download permutation and combination tricks and shortcuts pdf for bank exams

Permutation and Combination Tricks and Shortcuts For Bank Exams

Permutation and combination tricks are provided in this permutation and combination tricks for bank exams article. If you want to score full marks in the quantitative aptitude section of the bank exams you must be aware of the permutation and combination tricks. You can go through this permutation and combination tricks for bank exams article to be proficient with the permutation and combination tricks for bank exams. Here you will get to know the important permutation and combination tricks by downloading the permutation and combination tricks pdf which will be very useful for our future bank exams and other competitive exams. 

  • Factorial of a natural number n.

n! = 1 × 2 × 3 × 4 × .......× n

  • Permutation Formula for Bank Exams:

nPr = (n!) / (n-r)!

  • Combination Formula for Bank Exams:

nCr = (n!) / r!(n-r)!

  • The relationship between permutation and combination for r things taken from n things

nPr = r! × nCr

  • Practice various types of permutation and combination questions on a regular basis using these permutation and combination tricks for bank exams will help you to solve more questions within a short period of time with speed and accuracy. 
  • Try to solve different types of permutation and combination questions by using the permutation and combination tricks for bank exams and permutation and combination formulas for bank exams. It will help you to be familiarized with the permutation and combination topic. You can download the permutation and combination formula pdf for bank exams which is attached to this permutation and combination shortcuts pdf for bank exams article. 

Download permutation and combination formula pdf for bank exam

Permutation and Combination Tricks and Shortcuts For Bank Exams FAQs

Q. What is permutation formula in bank exams?

A. Permutation Formula for Bank Exams:

Q. What are the important permutation and combination tricks and shortcuts for bank exams?

A. The important permutation and combination tricks and shortcuts for bank exams are mentioned in the above article. Kindly go through the above permutation and combination tricks for bank exams post to know more permutation and combination tricks for the upcoming bank exams.

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50 + Quantitative Aptitude Questions on Permutation and Combination

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  • Feb 17, 2024

permutation and combination question bank pdf

Permutations and combinations in quantitative aptitude are two closely related topics that you should thoroughly prepare for if you’re getting ready for competitive exams . Two key mathematical terms that are commonly assessed in quantitative aptitude tests are permutations and combinations. A combination is a selection of items from a given set in any order, whereas a permutation is an arrangement of objects from a given set in a defined order. It can be challenging to distinguish between the two, but it can be made simpler by knowing some quick tricks for permutation and combination issues. This topic has been covered in great length in this article. Let’s now concentrate on practice questions about combinations and permutations.

This Blog Includes:

What are aptitude questions on permutation and combination, relation between permutation and combination, difference between permutation and combination, aptitude questions on permutation and combination.

The methods of choosing certain items from a collection of objects to create subsets, either with or without replacement, are permutation and combination. It outlines the different configurations for a given set of data. Permutations are when we choose items or data from a particular group; combinations are when we express them in a particular order. These terms are important for understanding mathematics. Let’s try to understand these important terms using some examples:

Aptitude Questions on Permutation and Combination involve a lot of formulas. The most important formulas are:

Must Read: 45+ Questions of Letter and Symbol Series with Answers

Permutation and Combination both refer to selecting objects from a set but with different importance on the order of selection. Mathematically, permutation counts arrangements, while combination counts selections. A mathematical relationship between permutation and combination is given by:

Below is a table highlighting the fundamental differences between permutation and combination, helping in understanding their usage in arranging elements and forming groups without considering sequence.

1.How many different ways can the letters of the word “APPLE” be arranged?

2.In how many ways can 5 books be arranged on a shelf?

3.A committee of 5 people is to be formed from a group of 10 candidates. In how many ways can the committee be formed?

How many 3-digit numbers can be formed using the digits 1, 2, 3, and 4 without repetition?

In how many ways can a committee of 3 men and 4 women be formed from 7 men and 5 women?

How many different 4-letter code words can be formed using the letters in the word “APPLE”?

How many different permutations can be formed from the letters of the word “BANANA”?

How many different 2-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 without repetition?

In how many ways can a president, vice president, and treasurer be selected from a group of 10 people?

How many different ways can the letters of the word “COMPUTER” be arranged?

A committee of 3 people is to be formed from 6 men and 4 women. In how many ways can the committee be formed if at least one man must be on the committee?

How many different 5-digit numbers can be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition?

In how many ways can the letters of the word “SUCCESS” be arranged if the vowels must always come together?

How many different 3-digit numbers can be formed using the digits 2, 3, 4, and 5 without repetition?

In how many ways can a president, vice president, and secretary be selected from a group of 10 people?

How many different permutations can be formed from the letters of the word “CHESS”?

In how many ways can a president, vice president, treasurer, and secretary be selected from a group of 10 people?

How many different 4-letter code words can be formed using the letters in the word “TRAIN”?

A group of 5 people consists of 2 men and 3 women. In how many ways can a committee of 3 people be formed if there must be at least 1 man and 1 woman?

How many different 5-letter code words can be formed using the letters A, B, C, D, E if repetition of letters is not allowed?

In how many ways can a president, vice president, and treasurer be selected from a group of 12 people?

How many different permutations can be formed from the letters of the word “ORANGE”?

In how many ways can the letters of the word “MISSISSIPPI” be arranged?

How many different 4-letter code words can be formed using the letters in the word “SHEET”?

In how many ways can a committee of 4 people be formed from a group of 10 candidates?

How many different permutations can be formed from the letters of the word “SEQUENCE”?

In how many ways can the letters of the word “MACHINE” be arranged?

How many different 3-letter code words can be formed using the letters A, B, C, D, E if repetition of letters is not allowed?

In how many ways can a president, vice president, and secretary be selected from a group of 8 people?

How many different permutations can be formed from the letters of the word “STATISTICS”?

Answer Key for Above Questions

Related Blogs

Understand the problem, identify if it’s a permutation or combination problem, apply relevant formulas, and ensure clarity in counting elements

Permutation involves arranging elements in a particular order, while combination involves selecting elements without considering the order. Both are frequently tested in competitive exams to assess quantitative aptitude skills.

In quantitative aptitude, permutation refers to arrangements where the order matters, while combination refers to selections where the order doesn’t matter. They’re crucial for solving problems involving arrangements and selections.

Permutation questions involve arranging elements in different orders, while combination questions involve selecting elements without regard to order. Both types of questions are common in quantitative aptitude assessments.

This was all about ” Aptitude Questions on Permutation and Combination”. For more such informative blogs, check out our Study Material Section , or you can learn more about us by visiting our   Indian exams page.

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COMMENTS

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  2. 1000+ Permutation and Combination PDF (Questions & Solution with

    Permutation and Combination PDF : Permutation and Combination is one of the most important topic that comes under Banking (IBPS, SBI, RBI, SEBI, NABARD, LIC), SSC (CGL, CHSL, MTS, CPO, SI, JE), Railway (RRB NTPC, Grade D, ALP, JE, TC), Defence (UPSC CDS/NDA/NA, Police, Army, Navy, Airforce) & Teaching Exams.

  3. PDF MATH 106 Lecture 2 Permutations & Combinations

    6 Permutations - Order Matters The number of ways one can select 2 items from a set of 6, with order mattering, is called the number of permutations of 2 items selected from 6 6×5 = 30 = 6 P 2 Example: The final night of the Folklore Festival will feature 3 different bands. There are 7 bands to choose from. How many different programs are possible?

  4. PDF PART 1 MODULE 5 FACTORIALS, PERMUTATIONS AND COMBINATIONS n! n

    1. Choosing a subset of r elements from a set of n elements; 2. Arranging the chosen elements. and. Referring to EXAMPLE 1.5.6 above, Gomer is choosing and arranging a subset of 9 elements from a set of 20 elements, so we can get the answer quickly by using the permutation formula, letting n = 20 and r = 9.

  5. PDF 10.5 Permutations and Combinations

    Section 10.5 Permutations and Combinations 569 10.5 Permutations and Combinations EEssential Questionssential Question How can a tree diagram help you visualize the number of ways in which two or more events can occur? Reading a Tree Diagram Work with a partner. Two coins are fl ipped and the spinner is spun. The tree diagram shows the possible ...

  6. PDF PERMUTATIONS AND COMBINATIONS

    The first step in solving problems involving permutations or combinations is determining whether or not it is necessary to consider the order of the elements. The following two questions in Example M will help to distinguish between when to use permutations and when to use combinations. EXAMPLE M The school hiking club has 10 members.

  7. Permutation and Combination Questions with Answers: PDF

    By practicing with these questions, you can strengthen your grasp on the concepts and improve your problem-solving skills. Additionally, we offer a convenient permutation and combination questions PDF for free download, allowing you to access a variety of Permutation and combination Questions with Answers anytime, anywhere. Permutation and ...

  8. PDF Lecture 1.3: Permutations and combinations

    Consider a quiz with four true/false and three multiple choice questions, (a){(e). The number of possible ways to answer the quiz is 2 2 2 2 5 5 5 = 24 53 = 2000: Proposition ... M. Macauley (Clemson) Lecture 1.3: Permutations and combinations Discrete Mathematical Structures 6 / 6.

  9. 18.600 F2019 Lecture 1: Permutations and combinations

    pdf. 412 kB 18.600 F2019 Lecture 1: Permutations and combinations ... Permutations and combinations Download File DOWNLOAD. Course Info Instructor Prof. Scott Sheffield; Departments Mathematics; As Taught In Fall 2019 Level Undergraduate. Topics Mathematics. Discrete Mathematics. Probability and Statistics. Learning Resource Types ...

  10. PDF CMSC 250: Permutations and Combinations

    3 Permutations of a Fixed Set with Repeated Elements 3.1 Introduction and Example Suppose we wish to permute the letters in the word BABAR keeping in mind that the two As and the two Bs are indistinguishable. How many possibilities are there? This is a permutation question (order matters) but we have repeated elements.

  11. PDF Permutations and Combina Tions

    7 PERMUTATIONS AND COMBINATIONS body of discovery is mathematical in form because there is no other guidance we can have - DARWIN 7.1 Introduction Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with 10 digits from 0 to 9.

  12. Question Bank

    Question Bank - Permutation and Combination - IIT JEE - Toppr1 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. ;)

  13. Permutation and Combination Practice Questions

    5. How many can four digits be formed using the digits 0, 1, 2, and 3 (repetition is allowed)? A. 12 B. 24 C. 256 D. 192. 6. In a word jumble, there are 8 consonants and 5 vowels given. Find out in how many ways can we form a 5-letter word having three consonants and 2 vowels?

  14. Permutations and Combinations Questions (With Solutions)

    Permutations and Combinations Questions Permutations and Combinations questions are provided here, along with detailed explanations to make the students understand easily. Permutation and combinations contain a large number of applications in our daily life. Thus, it is essential to learn and practise the fundamentals of these concepts.

  15. 1000+ Permutation and Combination Question Bank Pdf

    1000+ Permutation and Combination Question Bank Pdf - 1 Home aptitude exercise-5 Question: 1 In how many different ways can the letters of the word 'GAMBLE' be arranged? (A) 120 (B) 240 (C) 720 (D) 840 View Answer Question: 2 In how many different ways can the letters of the word 'SMART' be arranged? (A) 60 (B) 120 (C) 180 (D) 240 View Answer

  16. Permutation and Combination Formula PDF, Questions With Example

    Nov 15 2023 Access Free PDFs Here Permutation and Combination Formula PDF for Bank Exam: The selection of specific things from a group of things to form sub groups with or without replacement is said to be Permutation and Combination. These are the various ways of arranging a certain group of things.

  17. Permutation and Combination Solution of Question Bank 2

    Permutation and Combination Solution of Question Bank 2 | PDF | Consonant | Mathematics Permutation and Combination Solution of Question Bank 2 - Free download as PDF File (.pdf), Text File (.txt) or read online for free.

  18. PDF combinations and permutations

    The first paper contains 10 questions of which a candidate must answer 6 questions. Find the number of ways these 6 questions can be picked. The second paper has two sections. Section A has 9 questions of which a candidate must answer 6 questions and section B has 6 questions of which a candidate must answer 4 questions.

  19. Permutation and Combination Questions for Bank PO

    A permutation is a list of data and the combination is used for a group of data. Questions from Permutation and combination are frequently asked in various competitive exams. The list of exams where such type of questions are asked includes bank exams such as IBPS PO, SBI PO, IBPS RRB, IBPS Clerk, IBPS SO as well as exams like LIC AAO, SSC CGL etc.

  20. 12.2: Permutations and Combinations

    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.

  21. JEE Previous Year Question Bank On Permutation and Combination

    Download PDF of JEE Main Previous Year Solved Questions on Permutations and Combinations Other Important Concepts in Probability Binomial Theorem Combinations Permutation JEE Main Maths Permutations and Combinations Previous Year Questions With Solutions Question 1: A letter lock consists of three rings, each marked with ten different letters.

  22. Permutation and Combination Tricks And Shortcuts PDF Bank Exams

    Permutation and Combination Tricks and Shortcuts For Bank Exams: Permutation and combination tricks pdf for bank exams is added in this permutation and combinations tricks for bank exams article. Candidates who are effectively preparing for their upcoming bank exams and other competitive exams like ssc chsl, ssc cgl, ssc mts, railways and more shall utilize this permutation and combination ...

  23. 50 + Quantitative Aptitude Questions on Permutation and Combination

    Combination is a technique for determining the number of different possible arrangements where the order of selection is not relevant. Aptitude Questions on Permutation and Combination involve a lot of formulas. The most important formulas are: nPr = n! / (n−r)! nCr = n! / (r! * (n−r)!) nPr = n! / (n−k)!