BankingCareers - India's Bank Careers Portal

Permutation & Combination Solved Questions For IBPS, SBI and Other Bank Exams - Page 4

You will find 19 problems in 5 pages..

Permutation & Combination Solved Questions (Page 4 of 5)

Question 1

In how many different ways can the letters of the word 'SERVING' be arranged?

a) 5040 b) 720 c) 120 d) 40320

Answer : a) 5040

Solution :

There are 7 different letters in the word 'SERVING'.
Therefore, the number of arrangements of the seven letters of the word = Number of all permutations of 7 letters, taken 7 at a time =
nPn = n(n - 1)(n - 2) ... (n - n + 1) = n!
Here, n = 7 then required number of ways = 7! = 5040.


Question 2

In how many different ways can any 4 letters of the word 'WORKING' be arranged?

a) 5040 b) 840 c) 24 d) 120

Answer : b) 840

Solution :

There are 7 different letters in the word 'WORKING'.
Therefore, the number of arrangements of any 4 out of seven letters of the word = Number of all permutations of 7 letters, taken 4 at a time =
nPr = n(n - 1)(n - 2) ... (n - r + 1)
Here, n = 7 and r = 4, then we have
7p4 = 7 x 6 x 5 x 4 = 840.
Hence, the required number of ways is 840.


Question 3

In how many different ways can the letters of the word 'ARRANGEMENT' be arranged?

a) 2494800 b) 4989600 c) 831600 d) none of these.

Answer : a) 2494800

Solution :

There are 11 different letters in the word 'ARRANGEMENT', in which A occurs 2 times, R occurs 2 times, n occurs 2 times and E occurs 2 times.
Therefore, the number of different arrangements of 11 letters of the word = n!/(p1!)(p2)!.....(pr!)
Here, n = 11, p1 = 2, p2 = 2, p3 = 2 and p4 = 2.
Then, 11!/2!2!2!2! = 11!/16 = 2494800.
Hence, the required number of ways is 2494800.


Question 4

In how many different ways can the letters of the word 'DIGEST' be arranged such that the vowels may appear in the even places?

a) 30 b) 720 c) 144 d) 24

Answer : c) 144

Solution :

There are 4 consonants and 2 vowels in the word DIGEST.
Out of 6 places, 3 places odd and 3 places are even.
2 vowels can arranged in 3 even places in 3p2 ways = 3 x 2 = 6 ways.
And then 4 consonants can be arranged in the remaining 4 places in 4p4 ways = 4! = 24 ways.
Hence, the required number of ways = 6 x 24 = 144.


Permutation & Combination Solved Questions (Page 4 of 5)

Score Well In SBI & IBPS, PO & Clerk Exams