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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

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10!

\frac{10!}{2}

9!

none of these

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detailed solution

Correct option is B

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 $= \frac{10!}{2}$

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