JEE Mathematics Permutations and Combinations MCQs Set 23

Mathematics Objective Questions and Answers: Permutation and Combination

Access targeted multiple-choice questions for Permutation and Combination designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Permutation and Combination MCQs with Answers

View or download the dedicated Permutation and Combination MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: The figures 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is
(a) 72
(b) 90
(c) 96
(d) 98
Show Answer & Explanation

Answer: (b) 90

Explanation:
1. First digit 6, 7 or 8: \( 3 x 4! = 72 \) numbers.
2. First digit 5 and second digit 6, 7 or 8: \( 3 x 3! = 18 \) numbers.
3. \( 72 + 18 = 90 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 2: The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places is
(a) 24
(b) 12
(c) 18
(d) 30
Show Answer & Explanation

Answer: (c) 18

Explanation:
1. Odd places 1, 3, 5, 7 take the odd digits 1, 1, 3, 3: \( 4!/(2! 2!) = 6 \) ways.
2. Even places 2, 4, 6 take 2, 2, 4: \( 3!/2! = 3 \) ways.
3. \( 6 x 3 = 18 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 3: On the occasion of Deepawali, each student of a class sends greeting cards to the others. If there are 20 students in the class, the total number of greeting cards exchanged is
(a) \({}^{20}C_2\)
(b) \(2\cdot{}^{20}P_2\)
(c) \(2\cdot{}^{20}C_2\)
(d) None of these
Show Answer & Explanation

Answer: (c) \(2\cdot{}^{20}C_2\)

Explanation:
1. Each ordered pair (sender, receiver) gives one card: \( 20P2 = 20 x 19 \).
2. \({}^{20}P_2=2\cdot{}^{20}C_2=380\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 4: To fill 12 vacancies there are 25 candidates of which five are reserved for scheduled caste candidates. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, then the number of ways in which the selection can be made is
(a) \({}^5C_3\times{}^{22}C_9\)
(b) \({}^{22}C_9-{}^5C_3\)
(c) \({}^{22}C_3+{}^5C_3\)
(d) None of these
Show Answer & Explanation

Answer: (a) \({}^5C_3\times{}^{22}C_9\)

Explanation:
1. Choose 3 of the 5 scheduled caste candidates: \( 5C3 \).
2. The other 9 vacancies are filled from the remaining 22 candidates: \( 22C9 \).
3. Total \( 5C3 x 22C9 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 5: The number of 3-letter words, with or without meaning, which can be formed out of the letters of the word NUMBER. Statement I: when repetition is not allowed the number is 120. Statement II: when repetition is allowed the number is 216. Which statements are correct?
(a) Both I and II
(b) Only I
(c) Only II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Both I and II

Explanation:
1. Without repetition: \( 6P3 = 6 x 5 x 4 = 120 \), so Statement I is true.
2. With repetition: \( 6 x 6 x 6 = 216 \), so Statement II is true.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 6: A polygon is such that Statement I: the number of diagonals of an n-sided polygon is nC2 - n. Statement II: a polygon that has 44 diagonals has 10 sides. Which statements are correct?
(a) Only I
(b) Only II
(c) Both I and II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Only I

Explanation:
1. Statement I is true: \( nC2 \) counts all segments joining vertices, \( n \) of them are sides.
2. For 44 diagonals: \( n(n - 3)/2 = 44 \) gives \( n = 11 \), not 10, so Statement II is false.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 7: A committee of 7 has to be formed from 9 boys and 4 girls. Statement I: it can be done in 504 ways when the committee consists of exactly 3 girls. Statement II: it can be done in 588 ways when the committee consists of at least 3 girls. Which statements are correct?
(a) Both I and II
(b) Only I
(c) Only II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Both I and II

Explanation:
1. Exactly 3 girls: \( 4C3 x 9C4 = 4 x 126 = 504 \), so I is true.
2. At least 3 girls: \( 504 + 4C4 x 9C3 = 504 + 84 = 588 \), so II is true.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 8: Consider the statements I: if some or all of n objects are taken at a time, the number of combinations is \( 2^n - 1 \). II: an arrangement in a definite order which can be made by taking some or all of a number of things is called a permutation. Which statements are correct?
(a) Both I and II
(b) Only I
(c) Only II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Both I and II

Explanation:
1. Choosing none of the \( n \) objects is excluded, so the combinations are \( 2^n - 1 \) and I is true.
2. II is the definition of a permutation and is true.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 9: Consider the statements I: if nPr = nP(r+1) and nCr = nC(r-1), then n and r are 3 and 2 respectively. II: from a class of 32 students, 4 are to be chosen for a competition; this can be done in 32C2 ways. Which statements are correct?
(a) Only I
(b) Only II
(c) Both I and II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Only I

Explanation:
1. \( nPr = nP(r+1) \) gives \( n - r = 1 \); \( nCr = nC(r-1) \) gives \( n = 2r - 1 \). So \( r = 2 \) and \( n = 3 \), and I is true.
2. Choosing 4 students from 32 gives \( 32C4 \), not \( 32C2 \), so II is false.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 10: Consider the statements I: if n is an even natural number, the greatest among nC0, nC1, ..., nCn is nC(n/2). II: if n is an odd natural number, the greatest among them is nC((n-1)/2) or nC((n+1)/2). Which statements are correct?
(a) Both I and II
(b) Only I
(c) Only II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Both I and II

Explanation:
1. The binomial coefficients increase up to the middle and then decrease.
2. For even \( n \) there is one middle term, \( nC(n/2) \). For odd \( n \) the two middle coefficients are equal, \( nC((n-1)/2) = nC((n+1)/2) \).
3. Both statements are true.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 11: Consider the statements I: the continued product of the first n natural numbers is called a permutation. II: the L.C.M. of 4!, 5! and 6! is 720. Which statements are correct?
(a) Only II
(b) Only I
(c) Both I and II
(d) Neither I nor II
Show Answer & Explanation

Answer: (a) Only II

Explanation:
1. The continued product of the first \( n \) natural numbers is \( n! \), called the factorial, so I is false.
2. \( 6! = 720 \) is a multiple of 4! and 5!, so the \( L.C.M \). is 720 and II is true.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 12: Using the digits 1, 2, 3, 4, 5, 6, 7 a number of 4 different digits is formed. Match: A. how many numbers are formed, B. how many are exactly divisible by 2, C. how many are exactly divisible by 25, D. how many are exactly divisible by 4, with 1. 840, 2. 200, 3. 360, 4. 40. The correct pairing is
(a) A-1, B-3, C-4, D-2
(b) A-1, B-3, C-2, D-4
(c) A-3, B-1, C-4, D-2
(d) A-4, B-2, C-3, D-1
Show Answer & Explanation

Answer: (a) A-1, B-3, C-4, D-2

Explanation:
1. A: \( 7P4 = 840 \). \( B \): last digit even (3 ways), others \( 6 x 5 x 4 \): 360.
2. \( C \): ends in 25 or 75 (2 ways), the other two places \( 5 x 4 \): 40.
3. \( D \): last two digits among 12, 16, 24, 32, 36, 52, 56, 64, 72, 76 (10 pairs) \( x 5 x 4 = 200 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 13: Match: A. value of n if (n+2)! = 2550 x n!, B. value of n if (n+1)! = 12 (n-1)!, C. value of x if 1/9! + 1/10! = x/11!, D. value of P(15, 3), E. value of n if P(5, 3) = P(n, 4), with 1. 5, 2. 121, 3. 2730, 4. 49, 5. 3. The correct pairing is
(a) A-4, B-5, C-2, D-3, E-1
(b) A-4, B-2, C-5, D-3, E-1
(c) A-1, B-3, C-5, D-2, E-4
(d) A-1, B-5, C-3, D-2, E-4
Show Answer & Explanation

Answer: (a) A-4, B-5, C-2, D-3, E-1

Explanation:
1. A: \( (n+2)(n+1) = 2550 \) gives \( n = 49 \). \( B \): \( (n+1)n = 12 \) gives \( n = 3 \).
2. \( C \): \( x = 11 x 10 + 11 = 121 \). \( D \): \( 15 x 14 x 13 = 2730 \).
3. \( E \): \( P(5,3) = 60 = n(n-1)(n-2)(n-3) \) gives \( n = 5 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 14: Match: A. if P(n,4) = 20 P(n,2), the value of n; B. if 5Pr = 2 x 6P(r-1), the value of r; C. if 5Pr = 6P(r-1), the value of r; D. the value of 8!/(6! x 2!); with 1. 28, 2. 4, 3. 7, 4. 3. The correct pairing is
(a) A-3, B-4, C-2, D-1
(b) A-4, B-3, C-2, D-1
(c) A-3, B-4, C-1, D-2
(d) A-3, B-2, C-4, D-1
Show Answer & Explanation

Answer: (a) A-3, B-4, C-2, D-1

Explanation:
1. A: \( (n-2)(n-3) = 20 \) gives \( n = 7 \). \( B \): \( (7-r)(6-r) = 12 \) gives \( r = 3 \).
2. \( C \): \( (7-r)(6-r) = 6 \) gives \( r = 4 \). \( D \): \( 8 x 7/2 = 28 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 15: Match: A. if nC8 = nC2, find nC2; B. determine n if 2nC3 : nC2 = 12 : 1; C. determine n if 2nC3 : nC3 = 11 : 1; D. if nC8 = nC6, find nC2; with 1. 5, 2. 91, 3. 6, 4. 45. The correct pairing is
(a) A-4, B-1, C-3, D-2
(b) A-4, B-3, C-1, D-2
(c) A-2, B-1, C-3, D-4
(d) A-4, B-1, C-2, D-3
Show Answer & Explanation

Answer: (a) A-4, B-1, C-3, D-2

Explanation:
1. A: \( n = 10, 10C2 = 45 \). \( B \): \( 4(2n-1)/3 = 12 \) gives \( n = 5 \).
2. \( C \): \( 4(2n-1)/(n-2) = 11 \) gives \( n = 6 \). \( D \): \( n = 14, 14C2 = 91 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 16: If \({}^nC_9={}^nC_8\), what is the value of \({}^nC_{17}\)?
(a) 0
(b) 1
(c) 3
(d) 17
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \(9+8=n\) gives \(n=17\).
2. \({}^{17}C_{17}=1\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 17: If \({}^{10}C_x={}^{10}C_{x+4}\), then the value of \(x\) is
(a) 5
(b) 4
(c) 2
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. \(x+(x+4)=10\) gives \(x=3\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 18: Let \(T_n\) denote the number of triangles which can be formed using the vertices of a regular polygon of \(n\) sides. If \(T_{n+1}-T_n=21\), then \(n\) equals
(a) 7
(b) 5
(c) 6
(d) 4
Show Answer & Explanation

Answer: (a) 7

Explanation:
1. \(T_n={}^nC_3\), and \({}^{n+1}C_3-{}^nC_3={}^nC_2\).
2. \({}^nC_2=21\) gives \(n=7\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 19: If \(\dfrac1{6!}+\dfrac1{7!}=\dfrac x{8!}\), then \(x=2^m\). The value of \(m\) is
(a) 2
(b) 4
(c) 6
(d) 5
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. \(x=\dfrac{8!}{6!}+\dfrac{8!}{7!}=56+8=64\).
2. \(64=2^6\), so \(m=6\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 20: The value of \(\dfrac{n!}{(n-r)!}\) when \(n=6\), \(r=2\) is \(5m\). The value of \(m\) is
(a) 2
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. \(\dfrac{6!}{4!}=30=5\times6\), so \(m=6\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 21: Find \(n\) if \({}^{n-1}P_3:{}^nP_4=1:9\).
(a) 9
(b) 6
(c) 8
(d) 2
Show Answer & Explanation

Answer: (a) 9

Explanation:
1. \(\dfrac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\dfrac1n\).
2. \(\dfrac1n=\dfrac19\) gives \(n=9\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 22: Determine \(n\) if \({}^{2n}C_3:{}^nC_2=12:1\).
(a) 3
(b) 4
(c) 1
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. \(\dfrac{2n(2n-1)(2n-2)/6}{n(n-1)/2}=\dfrac{4(2n-1)}3\).
2. \(\dfrac{4(2n-1)}3=12\) gives \(n=5\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 23: If \({}^{12}P_r={}^{11}P_6+6\cdot{}^{11}P_5\), then \(r\) is equal to
(a) 6
(b) 5
(c) 7
(d) 8
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. \({}^{11}P_6+6\cdot{}^{11}P_5={}^{12}P_6\) (the identity \({}^{n+1}P_r={}^nP_r+r\cdot{}^nP_{r-1}\) with \( n = 11 \) and \( r = 6 \)).
2. So \( r = 6 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Question 24: The value of \({}^8C_1-{}^8C_2+{}^8C_3-{}^8C_4+{}^8C_5-{}^8C_6+{}^8C_7-{}^8C_8\) is
(a) 0
(b) 1
(c) 70
(d) 256
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \((1-1)^8=0=1-{}^8C_1+{}^8C_2-\cdots+{}^8C_8\).
2. So the given expression, which is the negative of the rest, equals 1.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Free study material for Permutations and Combinations

Multiple Choice Questions (MCQs) for JEE Mathematics Permutation and Combination

About Permutation and Combination MCQs for JEE Mathematics

Test your conceptual understanding of Permutation and Combination with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Mathematics, these problem sets build accuracy and prepare students for objective exams.

How to Verify Your MCQ Answers

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Enhance Speed with Online MCQ Tests

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Mathematics Permutations and Combinations MCQs Set 23?

You can get most exhaustive JEE Mathematics Permutations and Combinations MCQs Set 23 for free on StudiesToday.com. These MCQs for JEE Mathematics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics JEE material?

Yes, our JEE Mathematics Permutations and Combinations MCQs Set 23 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in JEE exams?

By solving our JEE Mathematics Permutations and Combinations MCQs Set 23, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for JEE Mathematics Permutations and Combinations MCQs Set 23?

Yes, Mathematics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Mathematics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Mathematics Permutations and Combinations MCQs Set 23 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.