In the below word how many words are there in which R and W are at the end positions?

RAINBOW

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.

Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

The number of points of intersection of 37 lines is C237. But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting C213 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C211 points, we get only one point B.

 Hence the number of intersection points of the lines is C237-C213-C211 +2 = 535

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. 

(i) 1 lady out of 4 and 4 gentlemen out of 6 

(ii) 2 ladies out of 4 and 3 gentlemen out of 6 

(iii) 3 ladies out of 4 and 2 gentlemen out of 6 

(iv) 4 ladies out of 4 and 1 gentlemen out of 6 

In case I the number of ways = C14×C46 = 4 x 15 = 60 

In case II the number of ways = C24×C36 = 6 x 20 = 120 

In case III the number of ways = C34×C26 = 4 x 15 = 60

In case IV the number of ways = C44×C16 = 1 x 6 = 6 

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. 

Required number of permutations = 18 x (17!) x 2 = 2 x 18!

First letter can be posted in 4 letter boxes in 4 ways. Similarly second letter can be posted in 4 letter boxes in 4 ways and so on.

Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

 Number of 7 digit numbers = 7!3!×2! = 420

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

=6!3!×2! = 60

Hence the required number of 7 digits numbers = 420 - 60 = 360

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

There are total 9 places out of which 4 are even and rest 5 places are odd.

4 women can be arranged at 4 even places in 4! ways.

and 5 men can be placed in remaining 5 places in 5! ways.

Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880