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

Answer: (b) 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

Answer: (d) 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

Answer: (a) 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

Answer: (a) 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

Answer: (d) 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}

Answer: (a) 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

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

Answer: (d) 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

Answer: (c) 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

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

Answer: (d) 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

Answer: (b) (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!)

Answer: (d) 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!

Answer: (b) 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!

Answer: (c) 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

Answer: (b) 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

Answer: (a) 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

Answer: (c) 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

Answer: (a) 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

Answer: (c) n!
Given, 2 × P(n, n – 2)
= 2 × {n!/(n – (n – 2))}
= 2 × {n!/(n – n + 2)}
= 2 × (n!/2)
= n!
So, 2 × P(n, n – 2) = n!