In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:

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.

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.

'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.

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.

Let the numbers be x and y.

Then, xy = 9375 and	x	= 15.
	y

xy	=	9375
(x/y)		15

⟹ y² = 625.

⟹ y = 25.

⟹ x = 15y = (15 x 25) = 375.

∴ Sum of the numbers = x + y = 375 + 25 = 400.

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

Then, (10x + y) - (10y + x) = 36

⟹ 9(x - y) = 36

⟹ x - y = 4.