RejinpaulPlacement
Search

Permutation and Combination - Aptitude Questions and Answers - RejinpaulPlacement




Find us on Facebook



Follow us on Google+


Google+

Quick Links

1.
Evaluate: 30 ! / 28 !
Answer & Solution
Answer: b) 870

Solution:   We have, 30 ! / 28 ! = [30 * 29 * (28 !)] / 28 ! = 30 * 29 = 870
close
2.
Find the value of
(i) 60 P3
(ii) 4 P4
Answer & Solution
Answer: b) 205320 , 24

Solution:   (i) 60 P3 = 60! / (60 - 3) ! = 60 ! /57 !
= [60 * 59 * 58 * (57 !)] / 57 ! = 60 * 59 * 58 = 205320

(ii) 4 P4 = 4! = 4 * 3 * 2 *`1) = 24
close
3.
Find the value of
(i) 10 C3
(ii) 100 C98
(iii) 50 C50
Answer & Solution
Answer: d) 120 , 4950 , 1

Solution:   (i) 10 C3 = [10 * 9 * 8] / 3! = [10 * 9 * 8] / [3 * 2 * 1] = 120

(ii) 100 C98 = 100 C(100 - 98) = 100 C2 = [100 * 99] / [2 * 1] = 4950

(iii) 50 C50 = 1 {Since, n Cn = 1}
close
4.
How many words can be formed by using all letters of the word 'BIHAR'?
Answer & Solution
Answer: c) 120

Solution:  The word BIHAR contains 5 different letters.

Require num,ber of words = 5 P5 = 5! = (5 * 4 * 3 * 2 * 1) = 120.
close
5.
How many words can be formed by using all the letters of the word 'DAUGHTER' so that vowels always come together?
Answer & Solution
Answer: c) 4320

Solution:   Give word contaons 8 different letters.

When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are DGHTR (AUE).

These 6 letters can be arranged in 6 P6 = 6! = 720 ways.

The vowels in the group (AUE) may be arranged in 3! = 6 ways.

Required number of words = (720 * 6) = 4320
close
6.
How many words can be formed from the letters of the word 'EXTRA', so that the vowels are never together?
Answer & Solution
Answer: b) 72

Solution:   The given word contains 5 different letters.

Taking vowels EA together, we treat them as one letter.
Then, the letters to be arranged are XTR (EA).

These letters can be arranged in 4! = 24 ways.

The vowels EA may be arranged amongest themselves in 2! = 2 ways.

Number of words , each having vowels together = (24 * 2) = 48

Total number of words formed by using all the letters of the given words
= 5! = (5 * 4 * 3 * 2 * 1) = 120.

Number of words, each having vowels never together = (120 - 48) = 72.
close
7.
How many words can be formed from the letters of the word 'DIRECTOR' so that the vowels are always together?
Answer & Solution
Answer: d) 2160

Solution:   In the given word, we treat the vowels IEO as one letter.

Thus, we have DRCTR (IEO).

This group has 6 letters of which R occurs 2 times and others are different.

Number of ways of arranging these letters = (6 !) / (2 !) = 360.

Now 3 vowels can be arranged among themselves in 3! = 6 ways.

Required number of ways = (360 * 6) = 2160.
close