Question

The number of ways in which $10$ candidates $A_1,A_2,A_3,\dots ,A_{10}$ can be ranked so that $A_1$ is always above $A_2,$ is

• A. $10!$
• B. $\frac{10!}{2}$
• C. $9!$
• D. None of these

Answer 1

The correct option is B $\frac{10!}{2}$

Ten candidates can be ranked in $10!$ ways.
In half of these ways, $A_1$ is above $A_2$ and in another half, $A_2$ is above $A_1.$
So, required number of ways is $\frac{10!}{2}$