# Class 11 Maths MCQ – Permutations and Combinations

This set of Class 11 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Permutations and Combinations”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. Permutation is also known as selection.
a) True
b) False

Explanation: Permutation is known as arrangement. Selection is another name for combinations.
It involves arrangement of letters, numbers, persons etc.

2. nPr = ________________
a) n!
b) $$\frac{n!}{r!}$$
c) $$\frac{n!}{(n-r)!}$$
d) $$\frac{n!}{(n-r)! r!}$$

Explanation: Permutation is known as arrangement. nPr means arranging r objects out of n.
nPr = $$\frac{n!}{(n-r)!}$$.

3. 6! = _____________
a) 24
b) 120
c) 720
d) 8

Explanation: We know, n! = n.(n-1).(n-2).(n-3)…..
6! = 6.5.4.3.2.1 = 720.

4. $$\frac{7!}{5!}$$ = ____________________
a) 7
b) 42
c) 230
d) 30

Explanation: We know, n! = n.(n-1).(n-2).(n-3)…… = n(n-1)!
$$\frac{7!}{5!} = \frac{7.6!}{5!} = \frac{7.6.5!}{5!}$$ = 7.6 = 42.

5. $$\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}$$. Find x.
a) 1
b) 2
c) 3
d) 4

Explanation: $$\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}$$.
We know, n! = n.(n-1). (n-2). (n-3) …………… = n(n-1)!
$$\frac{100}{10.9.8!} = \frac{1}{8!} + \frac{x}{9.8!}$$
=> $$\frac{100}{10.9} = \frac{1}{1} + \frac{x}{9}$$
=> $$\frac{10}{9} = 1 + \frac{x}{9}$$
=> $$\frac{x}{9} = \frac{10}{9} – 1 = \frac{1}{9}$$
=> x=1.
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6. nP0 = ________________
a) n!
b) 1
c) $$\frac{1}{(n)!}$$
d) (n-1)!

Explanation: We know, nPr = $$\frac{n!}{(n-r)!}$$.
nP0 = $$\frac{n!}{(n-0)!} = \frac{n!}{(n)!}$$ = 1.

7. nPn = ________________
a) n!
b) 1
c) $$\frac{1}{(n)!}$$
d) (n-1)!

Explanation: We know, nPr = $$\frac{n!}{(n-r)!}$$.
nPn = $$\frac{n!}{(n-n)!} = \frac{n!}{(0)!}$$ = n!

8. The number of permutations of n different objects taken r at a time, where repetition is allowed is _______________
a) n!
b) r!
c) nPr
d) nr

Explanation:
The number of permutations of n different objects taken r at a time, where repetition is allowed is n*n*n*n*n……………. r times = nr.

9. Find the number of permutations of word DEPENDENT.
a) 132400
b) 1512500
c) 1663200
d) 1723400

Explanation: There are total 9 letters out of which 1T, 2N, 2D, 3E, 1P.
Total number of permutations are $$\frac{9!}{3!2!2!} = \frac{9.8.7.6.5.4.3.2.1}{6*2*2} = \frac{362880}{24}$$ = 15120.

10. Find the number of 5 letter words which can be formed from word IMAGE without repetition using permutations.
a) 20
b) 60
c) 120
d) 240

Explanation: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE without repetition. So, total permutations are nPr = 5P5 = 5! = 5.4.3.2.1 = 120.

11. Find the number of 5 letter words that can be formed from word IMAGE using permutations if repetition is allowed.
a) 25
b) 120
c) 125
d) 3125

Explanation: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE with repetition allowed. So, total permutations are nr = 55 = 3125.

More MCQs on Class 11 Maths Chapter 7:

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