The number of ways in which 10 candidates A1,A2,…,A10 can be ranked so that A1 is always above A2,  is

# 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

1. A

$10!$

2. B

$\frac{10!}{2}$

3. C

$9!$

4. D

none of these

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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 $=\frac{10!}{2}$