How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

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.

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.

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.

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.