Aptitude - Permutation and Combination

Answer: Option B

Explanation:

In order that two books on Hindi are never together, we must place all these books as under:

X E X E X E X .... X E X

    where E - denotes the position of an English book and X that of a Hindi book.

Since there are 21 books on English, the number of places marked X are therefore, 22.

Now, 19 places out of 22 can be chosen in 22C19 = 22C3 =	22 x 21 x 20	= 1540 ways.
	3 x 2 x 1

Hence, the required number of ways = 1540.

Answer: Option D

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways	= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
	= 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1
	= 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1
	= (525 + 210 + 21)
	= 756.

Answer: Option C

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Answer: Option D

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =	7!	= 2520.
	2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in	5!	= 20 ways.
	3!

Required number of ways = (2520 x 20) = 50400.

Answer: Option C

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

	= (7C3 x 4C2)
	= 7 x 6 x 5 x 4 x 3 3 x 2 x 1 2 x 1
	= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves	= 5!
	= 5 x 4 x 3 x 2 x 1
	= 120.

Required number of ways = (210 x 120) = 25200.