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

RAINBOW

There are 12 - 3 = 9 patients.They can be seen in 12P9 = 79833600 ways

He can arrange his schedule in 8P6 = 20160 ways

The number of arrangements of 4 different digits taken 4 at a time is given by 4P4 = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

Number of permutations = 4P3 = 24

13P3= 13!/10! = 1716

This is the number of permutations of five things taken all at a time.

Therefore, answer = 5P5 = 120 ways

Since the word hexagon contains 7 different letters,the number of permutations is 7P3 = 7 x 6 x 5 =210

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 212. Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.

Therefore,there are 212- 1 = 4095 ways of selecting one or more candies