# 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

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.

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

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

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.

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

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

Explanation: nP3 = 4*nP2
$$\frac{n!}{(n-3)!} = 4 * \frac{n!}{(n-2)!}$$
=> n-2 = 4
=> n=6.

6. 4Pr = 4*5Pr-1. Find r.
a) 1
b) 2
c) 3
d) 4

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

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.

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

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

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.

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 