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

There are 6 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 4, marked 1, 3, 5.

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

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = ³P₃ = 3! = 6.

Total number of ways = (6 x 6) = 36.

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.

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.

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

∴ Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
	= ❨ 3 x 6 x 5 ❩ + ❨ 3 x 2 x 6 ❩ + 1 2 x 1 2 x 1
	= (45 + 18 + 1)
	= 64.

The word 'OPTICAL' contains 7 different letters.

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

Then, we have to arrange the letters PTCL (OIA).

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

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

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

Required number of ways	= (8C5 x 10C6)
	= (8C3 x 10C4)
	= ❨ 8 x 7 x 6 x 10 x 9 x 8 x 7 ❩ 3 x 2 x 1 4 x 3 x 2 x 1
	= 11760.

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.

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.

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

∴ Required number of numbers = (1 x 5 x 4) = 20.

Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =	❨	7 x 6	x 3	❩	= 63.
		2 x 1

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.