Question

If $n$ 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 $10!11!$

$n$ ! $(21-n)!=21$ ! $\frac{n!(21-n)!}{21!}=\frac{\left(21\right)!}{21C_n}$
For minimum value of, $\frac{\left(21\right)!}{21C_n}$, $n$ should be $10$
$\therefore \frac{\left(21\right)!}{{}^{21}{C}_{10}}=\frac{21!}{21!}\times 10!\times 11!=10!11!$