# If n is an integer between 0 and 21 then the minimum value of n! and (21- n)! is attained for n equals to

Multiply and divide n!(21-n)! by 21!, you get n!(21-n)! = $\frac{21!}{^{21}C_n}$
we know the max of ${^{n}C_r}$ is for r = n-1/2 if n is odd
therefore for n= 10 or 11 {both same value}, ${^{21}C_n}$ is maximum ,hence given term is minimum