Question

If ${}^{\prime }{n}^{\prime }$ is an integer between $0$ and $21$, then the minimum value of $n!(21-n)!$ is

• A. $9!2!$
• B. $10!11!$
• C. $20!$
• D. $21!$

Answer 1

The correct option is B $10!11!$

To find the minimum value of $n!(21-n)!=21!\frac{n!(21-n)!}{21!}=21!\frac{1}{{}^{21}{C}_n}=\frac{21!}{{}^{21}{C}_n}$

For minimum value, ${}^{21}{C}_n$ is maximum.

Maximum value of ${}^{21}{C}_n=\frac{{}^{21}{C}_{21-1}}{2}{=}^{21}{C}_{10}$

$\therefore$ Minimum value $=\frac{21!}{{}^{21}{C}_{10}}=\frac{21!}{21!}\times 10!11!=10!11!$