Mathematics Questions and Answers – Permutations-1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Permutations-1”.

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.

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