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

Solution:

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 $=\frac{10!}{2}$