In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

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

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.

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.

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

∴ (11x + 2)(2x + 2) = 144

⟹ 22x² + 26x - 140 = 0

⟹ 11x² + 13x - 70 = 0

⟹ (x - 2)(11x + 35) = 0

⟹ x = 2.

Hence, required number = 11x + 2 = 24.