Question

If $n$ is an integer lying between $0$ and $21$, then the least value of $n!(21-n)!$ is

• A. $1!20!$
• B. $11!10!$
• C. $9!12!$
• D. $2!19!$

Answer 1

Answer: (B)

$11!10!$

In pascals triangle middle terms has the highest value.
Therefore consider ${}^{21}{C}_n$
$=\frac{21!}{(21-n)!n!}$
For $(21-n)!n!$ to be least ${}^{21}{C}_n$ has to be maximum.
Therefore, since $21$ is odd we have two middle terms $T_{11}$ and $T_{12}$.
Hence for $n=11$,
$(21-n)!n!=11!\left(10\right)!$ which is less than $12!(9!)$ for $n=12$
The answer is B.