If ‘n’ is an integer between 0 and 21, then the minimum value of _{$n!\left(21-n\right)!$}  is

_{$n!\left(21-n\right)!=21!.\frac{n!.\left(21-n\right)!}{21!}=21!.\frac{1}{{}^{21}{C}_n}=\frac{21!}{{}^{21}{C}_n}$} For minimum value _{${}^{21}{C}_n$}  will be maximum. Maximum value of _{${}^{21}{C}_n={}^{21}{C}_{10}$  $\therefore$}  Maximum value _{$=\frac{21!}{{}^{21}{C}_{10}}=\frac{21!}{21!}\times 10!11!=10!11!$}