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
View Answer

Answer: b
Explanation: Permutation is known as arrangement. Selection is another name for combinations.
It involves arrangement of letters, numbers, persons etc.
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2. nPr = ________________
a) n!
b) \(\frac{n!}{r!}\)
c) \(\frac{n!}{(n-r)!}\)
d) \(\frac{n!}{(n-r)! r!}\)
View Answer

Answer: c
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
View Answer

Answer: c
Explanation: We know, n! = n.(n-1).(n-2).(n-3)…..
6! = 6.5.4.3.2.1 = 720.
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4. \(\frac{7!}{5!}\) = ____________________
a) 7
b) 42
c) 230
d) 30
View Answer

Answer: b
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
View Answer

Answer: a
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)!
View Answer

Answer: b
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)!
View Answer

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

Answer: d
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.
mathematics-questions-answers-permutations-1-q8

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

Answer: c
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.
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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
View Answer

Answer: c
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
View Answer

Answer: d
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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter