Question

The number of arrangements of $A_1,A_2....,A_{10}$ in a line so that $A_1$ is always above than $A_2$, is

• A. $2\times 10$
• B. $\frac{1}{2}\times 10$
• C. ${}^{10}{P}_2$
• D. ${}^{10}{C}_2$

Answer 1

The correct option is B $\frac{1}{2}\times 10$

Total number of arrangement in which $A_1$ is always above than $A_2=\frac{1}{2}(10!)$