In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Let original length = x metres and original breadth = y metres.

Original area = (xy) m².

New length =	❨	120	x	❩m	=	❨	6	x	❩m.
		100					5

New breadth =	❨	120	y	❩m	=	❨	6	y	❩m.
		100					5

New Area =	❨	6	x x	6	y	❩m2	=	❨	36	xy	❩m2.
		5		5					25

The difference between the original area = xy and new-area 36/25 xy is

= (36/25)xy - xy

= xy(36/25 - 1)

= xy(11/25) or (11/25)xy

∴ Increase % =	❨	11	xy x	1	x 100	❩%	= 44%.
		25		xy

Area to be plastered	= [2(l + b) x h] + (l x b)
	= {[2(25 + 12) x 6] + (25 x 12)} m2
	= (444 + 300) m2
	= 744 m2.

∴ Cost of plastering = Rs.	❨	744 x	75	❩	= Rs. 558.
			100

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.

Solving the two equations, we get: l = 63 and b = 40.

∴ Area = (l x b) = (63 x 40) m² = 2520 m².

2(l + b)	=	5
b		1

⟹ 2l + 2b = 5b

⟹ 3b = 2l

b =	2	l
	3

Then, Area = 216 cm²

⟹ l x b = 216

⟹l x	2	l	= 216
	3

⟹ l² = 324

⟹ l = 18 cm.

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm².

∴ Required number of tiles =	❨	1517 x 902	❩	= 814.
		41 x 41

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

∴ Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
	= ❨ 3 x 6 x 5 ❩ + ❨ 3 x 2 x 6 ❩ + 1 2 x 1 2 x 1
	= (45 + 18 + 1)
	= 64.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

	= (7C3 x 4C2)
	= ❨ 7 x 6 x 5 x 4 x 3 ❩ 3 x 2 x 1 2 x 1
	= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves	= 5!
	= 5 x 4 x 3 x 2 x 1
	= 120.

∴ Required number of ways = (210 x 120) = 25200.

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

∴ Required number of ways =	6!	= 360.
	(1!)(2!)(1!)(1!)(1!)

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.

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.