Permutation and Combination - General Questions (2)

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways	= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
	= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
	= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5 2 x 1 2 x 1 3 x 2 x 1 2 x 1
	= (24 + 90 + 80 + 15)
	= 209.

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways =	6!	= 360.
	(1!)(2!)(1!)(1!)(1!)

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.

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

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

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

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

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

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 (4 + 1 = 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.

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.