From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

'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.

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.

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.

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.

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.

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352.

Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.

Hence, required number = 253.

Let the ten's digit be x and unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

Let the ten's and unit digit be x and	8	respectively.
	x

Then,	❨	10x +	8	❩	+ 18 = 10 x	8	+ x
			x			x

⟹ 10x² + 8 + 18x = 80 + x²

⟹ 9x² + 18x - 72 = 0

⟹ x² + 2x - 8 = 0

⟹ (x + 4)(x - 2) = 0

⟹ x = 2.