Question

If $n$ is an integer with $0\le n\le 11$, then the minimum value of $n!(11-1)!$ is attained when a value of $n$ equals to

• A. 11
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (B)

We know, ${}^{11}{C}_n$ is maximum when $n=5$.
$\therefore {}^{11}{C}_n=\frac{11!}{n!(11-n)!}$
$\therefore n!(11-n)!$ is minimum when $n=5$.