Mathematics Questions and Answers – Permutations-2

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This set of Mathematics Questions and Answers for Aptitude test focuses on “Permutations-2”.

1. Find the number of 4 letter words which can be formed from word IMAGE using permutations without repetition.
a) 20
b) 60
c) 120
d) 240
View Answer

Answer: c
Explanation: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total permutations are nPr = 5P4 = 5! / (5-4)! = 5! = 5.4.3.2.1 = 120.
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2. Find the number of 4 letter words which can be formed from word IMAGE if repetition is allowed.
a) 120
b) 125
c) 625
d) 3125
View Answer

Answer: c
Explanation: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total permutations are nr = 54 = 625.

3. How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?
a) 504
b) 729
c) 1000
d) 720
View Answer

Answer: a
Explanation: 1-9 digits which means 9 digits are possible. We have to arrange 3 digits at a time out of 9 digits without repetition. So, total permutations are nPr = 9P3 = \(\frac{9!}{(9-3)!} = \frac{9!}{6!}\) = 9.8.7 = 504.
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4. How many 3-digit numbers are possible using permutations with repetition allowed if digits are 1-9?
a) 504
b) 729
c) 1000
d) 720
View Answer

Answer: b
Explanation: 1-9 digits which means 9 digits are possible. We have to arrange 3 digits at a time out of 9 digits with repetition allowed. So, total permutations are nr = 93 = 729.

5. If nP3 = 4*nP2. Find n.
a) 3
b) 2
c) 6
d) 5
View Answer

Answer: c
Explanation: nP3 = 4*nP2
\(\frac{n!}{(n-3)!} = 4 * \frac{n!}{(n-2)!}\)
=> n-2 = 4
=> n=6.
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6. 4Pr = 4*5Pr-1. Find r.
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: 4Pr = 4*5Pr-1
=> \(\frac{4!}{(4-r)!} = 4*\frac{5!}{(5-r+1)!}\)
=> \(\frac{(6-r)!}{(4-r)!} = 4*\frac{5!}{4!}\)
=> (6-r) (5-r) = 4*5
=> r=1.

7. Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels occur together.
a) 40320
b) 37440
c) 1440
d) 2880
View Answer

Answer: d
Explanation: There are 5 vowels in word EDUCATION. 5 vowels can be arranged in 5P5 i.e. 5! ways. When all vowels are together, 5 vowels together form one letter and remaining 3 letters i.e. together 4 letters can be arranged in 4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.
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8. Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels do not occur together.
a) 40320
b) 37440
c) 1440
d) 2880
View Answer

Answer: b
Explanation: There are 5 vowels in word EDUCATION. 5 vowels can be arranged in 5P5 i.e. 5! ways. When all vowels are together, 5 vowels together form one letter and remaining 3 letters i.e. together 4 letters can be arranged in 4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.
So, when all vowels do not occur together, total possible arrangements = 8! – 2880 = 40320 – 2880 = 37440.

9. In how many ways 2 red pens, 3 blue pens and 4 black pens can be arranged if same color pens are indistinguishable?
a) 362880
b) 1260
c) 24
d) 105680
View Answer

Answer: b
Explanation: Total number of pens are 2+3+4 = 9 out of which 2 are of 1 type, 3 are of 2nd type and 4 are of 3rd type so, total number of arrangements = \(\frac{9!}{2!3!4!} = \frac{9*8*7*6*5}{2*6}\) = 1260.
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10. Find the number of words which can be made using all the letters of the word IMAGE. If these words are written as in a dictionary, what will be the rank of MAGIE?
a) 97
b) 98
c) 99
d) 100
View Answer

Answer: c
Explanation: Words starting with letter A comes first in dictionary. Starting with A, number of words = 4! = 24. Starting with E, number of words = 4! = 24. Starting with I, number of words = 4! = 24. Starting with G, number of words = 4! = 24. Since our word also start with M so, we have to consider one more letter i.e. MA. Since our word also start with MA so, we have to consider one more letter i.e. MAE. Starting with MAE, number of words = 2! = 2. Since our word also start with MAG so, we have to consider one more letter i.e. MAGE. Starting with MAGE, only one letter i.e. MAGEI. After this, MAGIE comes. Total number of words before MAGIE = 24 + 24 + 24 + 24 + 2 = 98. So, rank of MAGIE is 99.

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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