Permutation & Combination Solved Questions For IBPS, SBI and Other Bank Exams - Page 5

You will find 19 problems in 5 pages..

Permutation & Combination Solved Questions (Page 5 of 5)

Question 1

In how many different ways can the letters of the word "CHARGES" be arranged in such a way that the vowels always come together?

a) 1440 b) 720 c) 360 d) 240

Answer : a) 1440

Solution :

The arrangement is made in such a way that the vowels always come together.
i.e., "CHRGS(AE)".

Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ways.

The vowels "AE" can be arranged themselves in 2! ways; i.e.,2! = 2 ways

Therefore, required number of ways = 720 x 2 = 1440 ways.


Question 2

In how many different ways can the letters of the word "COMPLAINT" be arranged in such a way that the vowels occupy only the odd positions?

a) 1440 b) 43200 c) 1440 d) 5420

Answer : b) 43200

Solution :

There are 9 different letters in the given word "COMPLAINT", out of which there are 3 vowels and 6 consonants.

Let us mark these positions as under:

[1] [2] [3] [4] [5] [6] [7] [8] [9]

Now, 3 vowels can be placed at any of the three places out of 5 marked 1, 3, 5, 7 and 9.

Number of ways of arranging the vowels = 5P3 = 5x4x3 = 60 ways.

Also, the 6 consonants at the remaining positions may be arranged in 6P6 ways = 6! ways = 720 ways.

Therefore, required number of ways = 60 x 720 = 43200 ways.


Question 3

In how many different ways can the letters of the word "CANDIDATE" be arranged in such a way that the vowels always come together?

a) 4320 b) 1440 c) 720 d) 840

Answer : a) 4320

Solution :

There are 9 letters in the given word, out of which 4 are vowels.

In the word "CANDIDATE" we treat the vowels "AIAE" as one letter.

Thus, we have CNDDT(AIAE).
Now, we have to arrange 6 letters, out of which D occurs twice.

Therefore, number of ways of arranging these letters = 6!/2! = 720 / 2 = 360 ways.

Now, AIAE has 4 letters, in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = 4!/2! = 1 x 2 x 3 x 4 / 2 = 12

Therefore, required number of words = (360 x 12) = 4320.


Question 4

In how many different ways can the letters of the word "RADIUS" be arranged in such a way that the vowels occupy only the odd positions?

a) 72 b) 144 c) 532 d) 36

Answer : d) 36

Solution :

There are 6 different letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

[1] [2] [3] [4] [5] [6]

Now, 3 vowels can be placed at any of the three places out of 3 marked 1, 3 and 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6 ways.

Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6 ways.

Therefore, total number of ways = 6 x 6 = 36.


Permutation & Combination Solved Questions (Page 5 of 5)

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