Question

In how many ways can $10$ examination papers be arranged so that the best and worst papers never come together?

Answer 1

Number of ways of arranging $10$ exam papers $={}^{10}{P}_{10}=10!\text{ways.}$

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 ${}^9{P}_9=9!=\text{ways.}$

Best and worst paper can themselves be arranged in ${}^2{P}_2=2!=2\text{ways.}$

Total number of arrangements with best and worst papers together $=9!\times 2\text{ways.}$

Hence, total number of arrangements with them not together $=10!-2\times 9!=8\times 9!\text{ways.}$