Multiple Choice Questions on Permutations and Combinations Class 11 OP Malhotra Exe-12G ISC Maths Solutions Ch-12. In this article you would learn to solve mcq problem on Permutations and Combinations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.

Permutations and Combinations Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-12

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-12	Permutations and Combinations
Writer	OP Malhotra
Exe-12(G)	Multiple Choice Questions.

Multiple Choice Questions on Permutations and Combinations

OP Malhotra ISC Class 11 Maths Solutions

Que-1: An office block has 7 entrance gates and 8 exit gates. In how many ways can a person enter and come out of the block?

(a) 15          (b) 50           (c) 57        (d) 56

Sol: (d) 56
Ways = 7 × 8 = 56

Que-2: There are 28 boys and 17 girls in a class. In how many ways can the class teacher select the class monitor if she can select either a boy or a girl?

(a) 45        (b) 11         (c) 17           (d) 46

Sol:  (a) 45
Total students = 28 + 17 = 45

Que-3: There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

(a) 3           (b) 36           (c) 66            (d) 108

Sol: (d) 108
Ways = ³C₂ × ⁹C₂ = 3 × 36 = 108

Que-4: The number of diagonals of a polygon of 20 sides is

(a) 210          (b) 190           (c) 180          (d) 170

Sol: (d) 170
Formula = n(n−3)/2
= 20(20−3)/2 = 20×17/2 = 170

Que-5: The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is

(a) 16C₁            (b) 16C₅             (c) 16C₉              (d) 20C₉

Sol: (c) 16C₉ 
22 − 2 − 4 = 16 remaining players.
Need 9 more players → 16C₉

Que-6: The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is

(a) 60          (b) 120            (c) 7200          (d) 720

Sol:  (c) 7200
Choose vowels = 4C2
Choose consonants = 5C3
Arrange 5 letters = 5!
Total = 4C2 × 5C3 × 5! = 7200

Que-7: How many words can be made from the letters of the word COMMITTEE?

(a) 9!            (b) 9! / 2!               (c) 9! / (2!)²           (d) 9! / (2!)³

Sol: (d) 9! / (2!)³)
Total letters in COMMITTEE = 9
Repeated letters: M(2), T(2), E(2)
Number of arrangements =
9! / (2! × 2! × 2!)
= 9! / (2!)³

Que-8: The number of permutations by taking all letters and keeping the vowels of the word COMBINE in the odd places is

(a) 96          (b) 144         (c) 512          (d) 576

Sol: (d) 576
Word: COMBINE → Total letters = 7
Vowels = O, I, E (3)
Odd positions = 1, 3, 5, 7
Ways to place vowels = 4P3 = 24
Arrange consonants (C, M, B, N) in remaining places = 4! = 24
Total permutations = 24 × 24 = 576

Que-9: If ⁴³C_r−6 = ⁴³C_3r+1, then the value of r is

(a) 12         (b) 8          (c) 6           (d) 10

Sol: (a) 12
Formula: n(n−3)/2 = 44
→ n = 11

Que-10: If a polygon has 44 diagonals, then the number of its sides is

(a) 7          (b) 8           (c) 10            (d) 11

Sol: (d) 11
Number of diagonals of a polygon = n(n − 3) / 2
Given: n(n − 3) / 2 = 44
n(n − 3) = 88
n² − 3n − 88 = 0
(n − 11)(n + 8) = 0
n = 11

Que-11: Everybody in a room shakes hands with everybody else. If the total number of handshakes is 66, then the total number of persons in the room is

(a) 11          (b) 12         (c) 13           (d) 14

Sol: (b) 12
Formula: Handshakes = n(n−1)/2
n(n−1)/2 = 66 ⇒ n = 12

Que-12: In an examination, there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all questions correct is

(a) 11          (b) 12          (c) 63           (d) 64

Sol:  (c) 63
Total ways = 4³ = 64
Correct answer only = 1
Fail ways = 64 − 1 = 63

Que-13: The number of permutations of the letters of the word CONSEQUENCE in which all the three E’s are together is

(a) 11! / 3!        (b) 9! / (2!2!)         (c) 11! / (2!2!)       (d) 9! / (2!2!3!)

Sol: (b) 9! / (2!2!)
Treat EEE as one letter → total letters = 9
Repeated letters: C(2), N(2)

Que-14: The number of ways in which 8 different keys can be put in a ring is

(a) 40320        (b) 4510         (c) 2520           (d) 5040

Sol: (c) 2520
Circular permutation = (n−1)!/2
= 7!/2 = 2520

Que-15: The number of triangles in a complete graph with 10 non-collinear vertices is

(a) 360            (b) 240           (c) 120            (d) 60

Sol: (c) 120
Number of triangles = C(10,3) = 120

Que-16: In an examination, a student has to answer 4 questions out of 5 questions, questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

(a) 20         (b) 3          (c) 5          (d) 9

Sol: (b) 3
Choose remaining 2 questions from 3 questions
Ways = C(3,1) = 3

Que-17: From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is

(a) at least 750 but less than 1000
(b) at least 1000
(c) less than 500
(d) at least 500 but less than 750

Sol:  (b) at least 1000
Choose novels = ⁶C₄ = 15
Choose dictionary = 3 ways
Arrange novels = 4! ways
Total = 15 × 3 × 4! = 1080 ≥ 1000

Que-18: The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together, is given by

(a) 30        (b) 5! × 4!         (c) 6! × 5!          (d) 7! × 5!

Sol: (c) 6! × 5!
Arrange men = (6−1)! = 5!
Women occupy spaces between men = 5! ways
Total = 6! × 5!

Que-19: A vehicle registration number consists of 2 letters of English alphabet followed by 4 digits where the first digit is not zero.
Then the total number of vehicles with distinct registration numbers is

(a) 26² × 10⁴
(b) ²⁶P₄ × ¹⁰P₄
(c) ²⁶P₄ × 9 × ¹⁰P₃
(d) 26² × 9 × 10³

Sol: (d) 26² × 9 × 10³)
Letters = 26²
First digit = 9 choices
Remaining digits = 10³
Total = 26² × 9 × 10³

Que-20: A candidate is required to answer 7 questions, out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. The number of ways in which he can choose the 7 questions is

(a) 780        (b) 640           (c) 820         (d) None of these

Sol: (a) 780
5 from group A and 2 from B = ⁶C₅ × ⁶C₂
4 from A and 3 from B = ⁶C₄ × ⁶C₃
3 from A and 4 from B = ⁶C₃ × ⁶C₄
2 from A and 5 from B = ⁶C₂ × ⁶C₅
Total = 780

Que-21: How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

(a) 8 × ⁶C₄ × ⁷C₄
(b) 6 × 7 × ⁸C₄
(c) 6 × 8 × ⁷C₄
(d) 7 × ⁶C₄ × ⁸C₄

Sol:  (d) 7 × ⁶C₄ × ⁸C₄)
Arrange other letters first then place S in gaps.
Total arrangements = 7 × ⁶C₄ × ⁸C₄

Que-22: Let T_n denote the number of triangles which can be formed using the vertices of a polygon of n sides.If T_n+1 − T_n = 21, then n =

(a) 5           (b) 7          (c) 6            (d) 4

Sol: (b) 7
T_n = ⁿC₃
T_n+1 − T_n = ⁿC₂ = 21
n(n−1)/2 = 21
n = 7

Que-23: For 2 ≤ r ≤ n,  ⁿC_r + ⁿC_r−1 + ⁿC_r−2 =

(a) ⁿ⁺¹C_r−1
(b) 2ⁿ⁺²C_r+1
(c) 2 × ⁿ⁻²C_r
(d) ⁿ⁺²C_r

 Sol: (d) ⁿ⁺²C_r)
ⁿC_r + ⁿC_r−1 + ⁿC_r−2
Using Pascal identity,
= ⁿ⁺²C_r

–: End Permutations and Combinations Class 11 OP Malhotra Exe-12G ISC Maths Ch-12 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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