Question

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

• A. $10!$
• B. $9!$
• C. $\frac{1}{2}9!$
• D. $\frac{1}{2}10!$

Answer 1

The correct option is D $\frac{1}{2}10!$

Total number of ways to arrange candidates $=10!$

$A_1$ and $A_2$ always be together and $A_1$ will be above $A_2$ and in the other half $A_2$ will be above $A_1.$

When $A_2$ in 2nd position $A_1$ can occupy only $1^{st}$.

When $A_2$ in 3rd position $A_1$ can occupy only $2^{nd}$.

So, the number of ways in witch $A_1$ and $A_2$ will arranged $=1+2+...+9=9.\frac{10}{2}$

In the remaining $8$ positions other students can be permuted in $8!$ ways.

Hence, total number of ways $=8!.9.\frac{10}{2}$ $=\frac{10}{2}!$

Option D is correct