Question

the number of ways in which candidates a1,a2,a3,..,a10 can be ranked if a1 is always above a2

Answer 1

Total number of ways to arrange a1 .. a10 = 10!
In 10! / 2 combinations a1 will be above a2 and in the other half a2 will be above a1.

Ans : 10! / 2

Answer 2

Answer:

10!/2will be the correct answer