Question

The number of ways in which ten candidates $A_1,A_2,...,A_{10}$ can be ranked, if $A_1$ is always above $A_{10}$ is?

• A. $5!$
• B. $2\left(5!\right)$
• C. $10!$
• D. $\frac{10!}{2}$

Answer 1

The correct option is D

$\frac{10!}{2}$

Explanation for correct option

Total number of ways in which $10$ candidates can be arranged $=10!$ ways

There are two cases: One in which $A_1$ is above $A_{10}$ and the other is $A_1$ is below $A_{10}$

Number of ways when $A_1$ is always above $A_{10}=\frac{10!}{2}$ ways

Hence, option (D) is correct