Question

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

• A. $9!2!$
• B. $9!$
• C. $\frac{10!}{2}$
• D. $10!$

Answer 1

The correct option is C $\frac{10!}{2}$

There are 10 candidates.

They can be arranged in $10!$ ways.

Out of these, there are $\frac{10!}{2}$ ways in which A1 is always ranked higher than A2 and similarly$\frac{10!}{2}$ ways in which A2 is ranked higher than A1.

So, according to given condition answer is $\frac{10!}{2}$.