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

Answer: (B)

$10!11!$

$n!(21-n)!=21!\frac{n!(21-n)!}{21!}=\frac{21!}{{{}^{21}c}_n}$

For minimum value, denominator should be maximum

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

So minimum value $\frac{21!}{{{}^{21}c}_{10}}=10!11!$