Answer to Question #281730 in Discrete Mathematics for Neha

Question

In how many ways can 10 different examination papers be scheduled so that

(i) the best and the worst always come together?

(ii) the best and the worst never come together?

Answer 1

(i) Let us consider the best and worst paper as one group and all other papers as different.

Number of ways of arranging 9 groups is $P(9,9)=9!$ ways.

Best and worst paper can themselves be arranged in $P(2,2)=2!$ ways.

Total number of arrangements with best and worst papers together is 

$$9!2!=725760$$

(ii) In how many ways can 10 examination papers be arranged without any condition?

$$P(10,10)=10!$$

Hence, total number of arrangements with the best and the worst not together is

$$10!-9!2!=9!(10-2)=8\cdot9!=2903040$$