Question

The number of ways in which 10 condidates `A_(1),A_(2),......,A_(10)` can be ranked so that `A_(1)` is always above `A_(2)`, is
A. 10!
B. `(10!)/(2)`
C. 9!
D. none of these

Answer 1

Ten candidates can be ranked in 10! ways. In half of these ways A1 is above `A_(2)` and in another half `A_(2)` is above `A_(1)`. So, required number of ways `=(10!)/(2)`