MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers

Permutations and Combinations Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Permutations and Combinations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 7 Quiz

Class 11 Maths Chapter 7 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Permutations and Combinations Class 11 Maths MCQ online test

Q1.	It is required to seat 5 men and 4 women in a row so that the women occupy the even places. The number of ways such arrangements are possible are
	A.8820
	B.2880
	C.2088
	D.2808

Ans: 2880
Total number of persons are 9 in which there are 5 men and 4 women
So total number of place = 9
Now women seat in even place
So total number of arrangement = 4! (_W_W_W_W_) (W-Woman)
Men sit in odd place
So total number of arrangement = 5! (MWMWMWMWM) (M-Man)
Now Total number of arrangement = 5! × 4! = 120 × 24 = 2880

Q2.	Six boys and six girls sit along a line alternately in x ways and along a circle (again alternatively in y ways), then
	A.x = y
	B.y = 12x
	C.x = 10y
	D.x = 12y

Ans: x = 12y
Given, six boys and six girls sit along a line alternately in x ways and along a circle
((again alternatively in y ways).
Now, x = 6! × 6! + 6! × 6!
⇒ x = 2 × (6!)2
and y = 5! × 6!
Now, x/y = {2 × (6!)2}/(5! × 6!)
⇒ x/y = {2 × 6! × 6! }/(5! × 6!)
⇒ x/y = {2 × 6!}/5!
⇒ x/y = {2 × 6 × 5!}/5!
⇒ x/y = 12
⇒ x = 12y

Q3.	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
	A.720
	B.420
	C.none of these
	D.5040

Ans: 720
The word LOGARITHMS has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= ¹⁰P₃
= 10 × 9 × 8
= 720

Q4.	A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of at least 3 girls
	A.588
	B.885
	C.858
	D.None of these

Ans: 588
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, the committee consists of atleast 3 girls:
⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃
= [{4! / (3! × 1!)} × {9! / (4! × 5!)}] + 9C₃
= [{(4 × 3!) /3!} × {(9 × 8 × 7 × 6 × 5!) / (4! × 5!)}] + 9! /(3! × 6!)
= [4 × {(9 × 8 × 7 × 6) / 4!}] + (9×8×7×6!)/(3! × 6!)
= [{4 × (9 × 8 × 7 × 6)} / (4 × 3 × 2 × 1)] + (9 × 8 × 7)/3!
= (9 × 8 × 7) + (9 × 8 × 7)/(3 × 2 × 1)
= 504 + (504/6)
= 504 + 84
= 588

Q5.	In how many ways can 12 people be divided into 3 groups where 4 persons must be there in each group?
	A.12!/{3! × (4!)³}
	B.12!/(4!)³
	C.Insufficient data
	D.none of these

Ans: 12!/{3! × (4!)³}
Number of ways in which
m × n”>
m × n distinct things can be divided equally into n
n”> groups
= (mn)!/{n! × (m!)n }
Given, 12(3 × 4) people needs to be divided into 3 groups where 4 persons must be there in each group.
So, the required number of ways = (12)!/{3! × (4!)n}

Q6.	How many factors are 2⁵ × 3⁶ × 5² are perfect squares
	A.24
	B.12
	C.16
	D.22

Ans: 24
Any factors of 2⁵ × 3⁶ × 5² which is a perfect square will be of the form 2^a × 3^b × 5^c
where a can be 0 or 2 or 4, So there are 3 ways
b can be 0 or 2 or 4 or 6, So there are 4 ways
a can be 0 or 2, So there are 2 ways
So, the required number of factors = 3 × 4 × 2 = 24

Q7.	If ⁿC₁₅ = ⁿC₆ then the value of ⁿC₂₁ is
	A.0
	B.1
	C.21
	D.None of these

Ans: 1
We know that
if ⁿC_r1 = ⁿC_r2
⇒ n = r₁ + r₂
Given, ⁿC₁₅ = ⁿC₆
⇒ n = 15 + 6
⇒ n = 21
Now, ²¹C₂₁ = 1

Q8.	If ⁿ⁺¹C₃ = 2 ⁿC₂, then the value of n is
	A.3
	B.4
	C.5
	D.6

Ans: 6
Given, ⁿ⁺¹C₃ = 2 ⁿC₂
⇒ [(n + 1)!/{(n + 1 – 3) × 3!}] = 2n!/{(n – 2) × 2!}
⇒ [{n × n!}/{(n – 2) × 3!}] = 2n!/{(n – 2) × 2}
⇒ n/3! = 1
⇒ n/6 = 1
⇒ n = 6

Q9.	There are 15 points in a plane, no two of which are in a straight line except 4, all of which are in a straight line. The number of triangle that can be formed by using these 15 points is
	A.¹⁵C₃
	B.490
	C.451
	D.415

Ans: 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

Q10.	In how many ways in which 8 students can be sated in a circle is
	A.40302
	B.40320
	C.5040
	D.50040

Ans: 5040
The number of ways in which 8 students can be sated in a circle = ( 8 – 1)!
= 7!
= 5040

Q11.	Let R = {a, b, c, d} and S = {1, 2, 3}, then the number of functions f, from R to S, which are onto is
	A.80
	B.16
	C.24
	D.36

Ans: 36
Total number of functions = 3⁴ = 81
All the four elements can be mapped to exactly one element in 3 ways, and exactly 3
elements in 3(2⁴ – 2) = 3(16 – 2) = 3 × 14 = 42
Thus the number of onto functions = 81 – 42 -3 = 81 – 45 = 36

Q12.	If (1 + x)ⁿ = C₀ + C₁ x + C₂ x² + …………..+ C_n xⁿ, then the value of C₀² + C₁² + C₂² + …………..+ C_nⁿ = ²ⁿC_n is
	A.(2n)!/(n!)
	B.(2n)!/(n! × n!)
	C.(2n)!/(n! × n!)2
	D.None of these

Ans: (2n)!/(n! × n!)
Given, (1 + x)ⁿ = C₀ + C₁ x + C₂ x² + ………….. + C_n xⁿ ………. 1
and (1 + x)ⁿ = C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ………….. C_r x^n-r + ………. + C_n-1 x + C_n ……….. 2
Multiply 1 and 2, we get
(1 + x)²ⁿ = (C₀ + C₁ x + C₂ x² + …………..+ C_n xⁿ) × (C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ………….. C_r x^n-r + ………. + C_n-1 x + C_n)
Now, equating the coefficient of xn on both side, we get
C₀² + C₁² + C₂² + …………..+ C_nⁿ = ²ⁿC_n = (2n)!/(n! × n!)

Q13.	The total number of 9 digit numbers of different digits is
	A.99!
	B.9!
	C.8 × 9!
	D.9 × 9!

Ans: 9 × 9!
Given digit in the number = 9
1st place can be filled = 9 ways = 9 (from 1-9 any number can be placed at first position)
2nd place can be filled = 9 ways (from 0-9 any number can be placed except the number which is placed at the first position)
3rd place can be filled = 8 ways
4th place can be filled = 7 ways
5th place can be filled = 6 ways
6th place can be filled = 5 ways
7th place can be filled = 4 ways
8th place can be filled = 3 ways
9th place can be filled = 2 ways
So total number of ways = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2
= 9 × 9!

Q14.	The number of ways in which 6 men add 5 women can dine at a round table, if no two women are to sit together, is given by
	A.30
	B.5 ! × 5 !
	C.5 ! × 4 !
	D.7 ! × 5 !

Ans: 5 ! × 5 !
Again, 6 girls can be arranged among themselves in 5! ways in a circle.
So, the number of arrangements where boys and girls sit attentively in a circle = 5! × 5!

Q15.	There are 15 points in a plane, no two of which are in a straight line except 4, all of which are in a straight line. The number of triangle that can be formed by using these 15 points is
	A.¹⁵C₃
	B.490
	C.451
	D.415

Ans: 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

Q16.	The number of 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated are
	A.110
	B.120
	C.130
	D.140

Ans: 120
A number is divisible by 10 if the unit digit of the number is 0.
Given digits are 0, 1, 3, 5, 7, 9
Now we fix digit 0 at unit place of the number.
Remaining 5 digits can be arranged in 5! ways
So, total 6-digit numbers which are divisible by 10 = 5! = 120

Q17.	6 men and 4 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is
	A.604800
	B.17280
	C.120960
	D.518400

Ans: 604800
6 men can be sit as
× M × M × M × M × M × M ×
Now, there are 7 spaces and 4 women can be sit as ⁷P₄ = ⁷P₃ = 7!/3! = (7 × 6 × 5 × 4 × 3!)/3!
= 7 × 6 × 5 × 4 = 840
Now, total number of arrangement = 6! × 840
= 720 × 840
= 604800

Q18.	A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of exactly 3 girls
	A.540
	B.405
	C.504
	D.None of these

Ans: 504
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, If in committee consist of exactly 3 girls:
⁴C₃ × ⁹C₄
= {4! / (3! × 1!)} × {9! / (4! × 5!)}
= {(4×3!) /3!} × {(9 × 8 × 7 × 6 × 5!) / (4! × 5!)}
= 4 × {(9 × 8 × 7 × 6) / 4!}
= {4 × (9 × 8 × 7 × 6)} / (4 × 3 × 2 × 1)
= 9 × 8 × 7
= 504