Question

If $n$ is an integer between $0$ and $21$, then the minimum value of $n!(21-n)!$ is attained for $n=$

• A. $1$
• B. $10$
• C. $12$
• D. $20$

Answer 1

Answer: (A)

$1$

Let us consider $(1+x{)}^{21}$
Hence $T_{r+1}={}^{21}{C}_r{x}^r$
We know that both the middle terms will have the highest value of binomial coefficient.
The middle terms in thew above binomial expansion are ${11}^{th}$ and ${12}^{th}$ terms.
$T_{10+1}={}^{21}{C}_{10}{x}^{10}=\frac{21!}{10!\left(11\right)!}{x}^{10}$

$T_{11+1}={}^{21}{C}_{11}{x}^{11}=\frac{21!}{11!\left(10\right)!}{x}^{11}$

Hence $n!(21-n)!$ is least for $n=0,n=21$ followed by $n=11$ and $n=10$