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

Explanation: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total permutations are

^{n}P

_{r}=

^{5}P

_{4}= 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

View Answer

Explanation: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total permutations are n

^{r}= 5

^{4}= 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

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

^{n}P

_{r}=

^{9}P

_{3}= \(\frac{9!}{(9-3)!} = \frac{9!}{6!}\) = 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

View Answer

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 n

^{r}= 9

^{3}= 729.

5. If ^{n}P_{3} = 4*^{n}P_{2}. Find n.

a) 3

b) 2

c) 6

d) 5

View Answer

Explanation:

^{n}P

_{3}= 4*

^{n}P

_{2}

\(\frac{n!}{(n-3)!} = 4 * \frac{n!}{(n-2)!}\)

=> n-2 = 4

=> n=6.

6. ^{4}P_{r} = 4*^{5}P_{r-1}. Find r.

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation:

^{4}P

_{r}= 4*

^{5}P

_{r-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

Explanation: There are 5 vowels in word EDUCATION. 5 vowels can be arranged in

^{5}P

_{5}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

^{4}P

_{4}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

View Answer

Explanation: There are 5 vowels in word EDUCATION. 5 vowels can be arranged in

^{5}P

_{5}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

^{4}P

_{4}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

Explanation: Total number of pens are 2+3+4 = 9 out of which 2 are of 1 type, 3 are of 2

^{nd}type and 4 are of 3

^{rd}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

View Answer

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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