Question

In how many ways can $10$ examination papers be arranged so that the best and worst papers are always together?

• A. $10!$
• B. $2\times 9!$
• C. $9!$
• D. none of these

Answer 1

Answer: (B)

$2\times 9!$

The remaning papers other than the best and worst $=8$

Totally $9$ sets/papers (or combined) are to be arranged

Total possiblities will be

$=9!\times 2!=2!\times 9!$