CBSE Class 10 Mathematics Permutations and Combinations MCQs Set 11

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Permutations and Combinations

Access targeted multiple-choice questions for Permutations and Combinations designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Permutations and Combinations MCQs for Class 10 Mathematics

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Multiple Choice Questions

Question 1: The number of ways in which 5 letters can be posted in 4 letter boxes is
(a) 5⁴
(b) 20
(c) 25
(d) 4⁵
Show Answer & Explanation

Answer: (d) 4⁵

Explanation:
1. Each letter can go into any of the 4 boxes independently: 4⁵ ways.

Teacher's Note:
1. Count the choices for each letter, not each box.

Question 2: The total number of numbers divisible by 5 and lying between 4000 and 5000 that can be formed from the digits 4, 5, 6, 7 and 8 (digits may repeat) is
(a) 125
(b) 120
(c) 25
(d) 625
Show Answer & Explanation

Answer: (c) 25

Explanation:
1. The first digit must be 4 and the last digit must be 5 (divisible by 5).
2. The two middle digits can be any of the 5 digits: 5 × 5 = 25.

Teacher's Note:
1. Use the multiplication principle: multiply the number of choices at each stage.

Question 3: The number of different signals that can be given using any number of flags from 5 flags of different colours (flags arranged in a row) is
(a) 325
(b) 240
(c) 60
(d) None of these
Show Answer & Explanation

Answer: (a) 325

Explanation:
1. Signals with r flags = ⁵Pᵣ: 5 + 20 + 60 + 120 + 120 = 325.

Teacher's Note:
1. Add the permutations for r = 1 to 5.

Question 4: The number of different ways a person can reach Mumbai from Ahmedabad if there are 4 different routes from Ahmedabad to Surat and 3 different routes from Surat to Mumbai is
(a) 7
(b) 12
(c) 24
(d) 36
Show Answer & Explanation

Answer: (b) 12

Explanation:
1. 4 × 3 = 12 ways.

Teacher's Note:
1. Use the multiplication principle: multiply the number of choices at each stage.

Question 5: The number of different ways 3 travellers can stay in 4 hotels, each in a different hotel, is
(a) 24
(b) 30
(c) 36
(d) 48
Show Answer & Explanation

Answer: (a) 24

Explanation:
1. The first has 4 choices, the second 3 and the third 2: 4 × 3 × 2 = 24.

Teacher's Note:
1. This is ⁴P₃.

Question 6: A letter lock contains 4 rings, each ring having 5 letters. If the lock opens in only one arrangement of 4 letters, the number of unsuccessful attempts possible is
(a) 120
(b) 625
(c) 624
(d) None of these
Show Answer & Explanation

Answer: (c) 624

Explanation:
1. Total arrangements = 5⁴ = 625. Only one opens the lock, so 624 fail.

Teacher's Note:
1. Subtract the single successful arrangement.

Question 7: There are 10 digits from 0 to 9. The number of 5-digit numbers that can be formed so that no two digits are the same is
(a) 12216
(b) 26127
(c) 62716
(d) 27216
Show Answer & Explanation

Answer: (d) 27216

Explanation:
1. The first digit has 9 choices (not 0), then 9, 8, 7 and 6 choices: 9 × 9 × 8 × 7 × 6 = 27216.

Teacher's Note:
1. The first digit cannot be 0.

Question 8: The number of 4-digit numbers that can be formed using the digits 0 to 9 (repetition allowed) is
(a) 4500
(b) 9000
(c) 10500
(d) 12000
Show Answer & Explanation

Answer: (b) 9000

Explanation:
1. 9 choices for the first digit and 10 for each of the other three: 9 × 10³ = 9000.

Teacher's Note:
1. The first digit cannot be 0.

Question 9: The number of different ways in which 5 different balls can be distributed among 3 different boxes, each box holding any number of balls, is
(a) 120
(b) 240
(c) 243
(d) 486
Show Answer & Explanation

Answer: (c) 243

Explanation:
1. Each ball can go into any of the 3 boxes: 3⁵ = 243.

Teacher's Note:
1. Choose a box for each ball.

Question 10: The value of r when ¹⁰Pᵣ = 720 is
(a) 3
(b) 5
(c) 7
(d) None of these
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. ¹⁰P₃ = 10 × 9 × 8 = 720, so r = 3.

Teacher's Note:
1. Multiply r descending factors from 10.

Question 11: The number of ways in which 3 scholarships of unequal value can be awarded to 17 candidates, if no candidate gets more than one scholarship, is
(a) ¹⁷P₁
(b) ¹⁷P₁₇
(c) ¹⁷P₃
(d) None of these
Show Answer & Explanation

Answer: (c) ¹⁷P₃

Explanation:
1. The scholarships are different, so order matters: 17 × 16 × 15 = ¹⁷P₃.

Teacher's Note:
1. Different prizes mean permutations, not combinations.

Question 12: The number of ways in which 17 billiard balls can be arranged in a row if 7 are black, 6 red and 4 white is
(a) ¹⁷P₇ × ¹⁷P₆ × ¹⁷P₄
(b) ¹⁷P₃
(c) 17!/(7! 6! 4!)
(d) ¹⁷C₇ × ¹⁷C₆ × ¹⁷C₄
Show Answer & Explanation

Answer: (c) 17!/(7! 6! 4!)

Explanation:
1. Balls of the same colour are identical, so divide 17! by 7! 6! 4!.

Teacher's Note:
1. Arrangements of identical objects: n!/(p! q! r!).

Question 13: Five friends A, B, C, D and E want a group photograph sitting in a single row. The number of distinct photographs is
(a) 60
(b) 120
(c) 240
(d) None of these
Show Answer & Explanation

Answer: (b) 120

Explanation:
1. 5! = 120.

Teacher's Note:
1. Arrange 5 distinct people in a row.

Question 14: Five friends A, B, C, D and E sit in a row for a photograph. In how many of these photographs would A be sitting in the middle?
(a) 12
(b) 18
(c) 24
(d) None of these
Show Answer & Explanation

Answer: (c) 24

Explanation:
1. Fix A in the middle and arrange the other 4: 4! = 24.

Teacher's Note:
1. Fix the special person first.

Question 15: Five friends A, B, C, D and E sit in a row. In how many photographs would A and B be sitting next to each other?
(a) 24
(b) 30
(c) 36
(d) 48
Show Answer & Explanation

Answer: (d) 48

Explanation:
1. Treat AB as one block: 4! arrangements of 4 units, times 2 orders inside the block = 48.

Teacher's Note:
1. Glue the pair together and multiply by 2.

Question 16: A group of 6 students comprising 3 boys and 3 girls is arranged in a straight line so that no two boys sit together. The number of ways is
(a) 36
(b) 72
(c) 108
(d) 144
Show Answer & Explanation

Answer: (d) 144

Explanation:
1. Arrange the 3 girls: 3! = 6 ways. This leaves 4 gaps; the boys occupy 3 of them in order: ⁴P₃ = 24.
2. Total = 6 × 24 = 144.

Teacher's Note:
1. Arrange the unrestricted group first, then slot the restricted group in the gaps.

Question 17: The number of ways to arrange the letters of the word FIGMENT when it starts with F is
(a) 360
(b) 720
(c) 1080
(d) 1440
Show Answer & Explanation

Answer: (b) 720

Explanation:
1. Fix F first and arrange the remaining 6 distinct letters: 6! = 720.

Teacher's Note:
1. Fix the first letter.

Question 18: The number of ways to arrange the letters of FIGMENT with vowels at the first and last positions is
(a) 120
(b) 240
(c) 360
(d) 720
Show Answer & Explanation

Answer: (b) 240

Explanation:
1. The vowels are I and E: they take the two end places in 2! ways.
2. The other 5 letters fill the middle in 5! = 120 ways: 2 × 120 = 240.

Teacher's Note:
1. Fill the restricted places first.

Question 19: The number of ways in which 7 boys can sit at a round table so that 2 particular boys are next to each other is
(a) 120
(b) 240
(c) 360
(d) 480
Show Answer & Explanation

Answer: (b) 240

Explanation:
1. Treat the two as one unit: 6 units in a circle give 5! arrangements, and the pair can swap: 2 × 120 = 240.

Teacher's Note:
1. Circular arrangements of n objects: (n − 1)!.

Question 20: The number of four-digit numbers in which all the four digits are distinct is
(a) 4536
(b) 3024
(c) 5040
(d) 9000
Show Answer & Explanation

Answer: (a) 4536

Explanation:
1. 9 × 9 × 8 × 7 = 4536 (the first digit cannot be 0).

Teacher's Note:
1. Choose digit by digit.

Question 21: How many six-digit numbers are there in which no digit is repeated, even digits appear at even places, odd digits appear at odd places, and the number is divisible by 4?
(a) 3600
(b) 2700
(c) 2160
(d) 1440
Show Answer & Explanation

Answer: (d) 1440

Explanation:
1. The last two digits form 10d₅ + d₆ with d₅ odd and d₆ even. 10 × odd leaves remainder 2 on division by 4, so d₆ must leave remainder 2: d₆ is 2 or 6.
2. d₆: 2 ways; d₅: 5 ways; the other odd places (positions 1, 3): 4 × 3 = 12 ways; the other even places (positions 2, 4): 4 × 3 = 12 ways.
3. Total = 2 × 5 × 12 × 12 = 1440.

Teacher's Note:
1. Divisibility by 4 depends only on the last two digits.

Question 22: If ⁿPᵣ = ⁿPᵣ₊₁ and ⁿCᵣ = ⁿCᵣ₋₁, then the values of n and r are
(a) n = 4, r = 2
(b) n = 4, r = 3
(c) n = 3, r = 2
(d) n = 3, r = 1
Show Answer & Explanation

Answer: (c) n = 3, r = 2

Explanation:
1. ⁿPᵣ = ⁿPᵣ₊₁ gives (n − r)! = (n − r − 1)!, so n − r = 1.
2. ⁿCᵣ = ⁿCᵣ₋₁ gives n = 2r − 1.
3. So r = 2 and n = 3.

Teacher's Note:
1. Use the factorial forms of nPr and nCr.

Question 23: In an examination a student has to answer 4 questions out of 5, questions 1 and 2 being compulsory. The number of ways in which the student can make a choice is
(a) 6
(b) 3
(c) 2
(d) 60
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Questions 1 and 2 are fixed. Choose 2 more from the remaining 3: ³C₂ = 3.

Teacher's Note:
1. Fix the compulsory items first.

Question 24: Out of 7 consonants and 4 vowels, the number of words that can be formed with 3 consonants and 2 vowels is
(a) 462
(b) 2800
(c) 25200
(d) 24540
Show Answer & Explanation

Answer: (c) 25200

Explanation:
1. Choose the letters: ⁷C₃ × ⁴C₂ = 35 × 6 = 210.
2. Arrange the 5 chosen letters: 5! = 120, so 210 × 120 = 25200.

Teacher's Note:
1. A word needs selection and then arrangement.

Question 25: If ⁿ⁺²C₈ : ⁿ⁻²P₄ = 57 : 16, the value of n is
(a) 16
(b) 17
(c) 18
(d) 19
Show Answer & Explanation

Answer: (d) 19

Explanation:
1. Writing both sides out for n = 19: ²¹C₈ = 203490 and ¹⁷P₄ = 57120, and 203490 : 57120 = 57 : 16 (203490 × 16 = 3255840 = 57 × 57120).
2. Smaller values fail, since n − 2 must be at least 4 for ⁿ⁻²P₄ to be non-zero.

Teacher's Note:
1. Test the options using the definition of nCr and nPr.

Question 26: The total number of squares in a square grid of order n (an n × n grid of small squares) is
(a) n(n + 1)/2
(b) n²(n + 1)²/4
(c) n(n + 1)(2n + 1)/6
(d) n(n + 1)(n + 2)/6
Show Answer & Explanation

Answer: (c) n(n + 1)(2n + 1)/6

Explanation:
1. Squares of side k number (n − k + 1)², so the total is Σ k² for k = 1..n = n(n + 1)(2n + 1)/6.

Teacher's Note:
1. For a 3 × 3 grid: 9 + 4 + 1 = 14.

Question 27: The number of rectangles formed by 6 horizontal and 4 vertical lines is n, and by 5 vertical and 5 horizontal lines is m. Then
(a) n = m
(b) n ≥ m + 1
(c) m ≤ n
(d) m > n + 5
Show Answer & Explanation

Answer: (d) m > n + 5

Explanation:
1. n = ⁶C₂ × ⁴C₂ = 15 × 6 = 90 and m = ⁵C₂ × ⁵C₂ = 100.
2. m = 100 > 95 = n + 5.

Teacher's Note:
1. A rectangle needs 2 horizontal and 2 vertical lines.

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Permutations and Combinations Objective Questions & Solutions for Class 10 Mathematics

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FAQs

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