Question

The number of ways in which ten candidates $A_{1},\mspace{6mu} A_{2},\mspace{6mu}.......A_{10}$ can be ranked such that $A_{1}$ is always above $A_{10}$ is.

• A. $5\mspace{6mu}!$
• B. $2(5\mspace{6mu}!)$
• C. $10\mspace{6mu}!$
• D. $\frac{1}{2}(10\mspace{6mu}!)$

Answer 1

Answer: (D)

$\frac{1}{2}(10\mspace{6mu}!)$

Answer 2

Without any restriction the 10 persons can be ranked among themselves in $10\mspace{6mu}!$ ways; but the number of ways in which $A_{1}$ is above $A_{10}$ and the number of ways in which $A_{10}$ is above $A_{1}$ make up $10\mspace{6mu}!$. Also the number of ways in which $A_{1}$ is above $A_{10}$ is exactly same as the number of ways in which $A_{10}$ is above $A_{1}$.

Therefore the required number of ways $= \frac{1}{2}(10\mspace{6mu}!)$.