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Permutation

The number of...

Question

The number of ways in which ten candidates $A_{1}, A_{2}, . . . . . . A_{10}$ can be ranked such that $A_{1}$ is always above $A_{10}$ is

5!

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2 (5!)

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10!

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\frac{1}{2} (10!)

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Solution

The correct option is C $\frac{1}{2} (10!)$
Total number of ways to arrange $A_{1}, A_{2} . . A_{10} = 10!$
In $\frac{10!}{2}$ combinations $A_{1}$ will be above $A_{2}$ and in the other half $A_{2}$ will be above $A_{1}$ .

Ans : $\frac{10!}{2}$

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