In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

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.

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.

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.

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.

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.

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.

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

∴ Number of ways of arranging these letters =	8!	= 10080.
	(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =	4!	= 12.
	2!

∴ Required number of words = (10080 x 12) = 120960.

'LOGARITHMS' contains 10 different letters.

Required number of words	= Number of arrangements of 10 letters, taking 4 at a time.
	= 10P4
	= (10 x 9 x 8 x 7)
	= 5040.

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.

Let the numbers be x and y.

Then, x + y = 25 and x - y = 13.

4xy = (x + y)² - (x- y)²

   = (25)² - (13)²

   = (625 - 169)

   = 456

∴ xy = 114.