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

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

There are 10 candidates to be arranged in 10 places

$\text{No. of ways in which}$ $A_1$ $\text{will always come before}$ $A_2$
$=$ no. of ways in which two places can be selected out of 10 places $={}^{10}{C}_2$

$\text{And remaining 8 students can be arranged on remaining 8 places}$ $=8!$ ways

$\text{Required arrangement}$ $={}^{10}{C}_2\times 8!=\frac{10!}{8!\times 2!}\times 8!=\frac{10!}{2}$