Question

If $n$ is a positive integer, then $\left(n^2\right)!$ is

• A. divisible by $2^n$ , but not by n!
• B. divisible by $n!$, but not by $\left(n^2\right)!$
• C. divisible by $\left(n^2\right)$, but not by $(n!{)}^n$
• D. divisible by $(n!{)}^n$

Answer 1

Answer: (D)

divisible by $(n!{)}^n$

Consider $n$ identical objects of first kind, $n$ identical objects of second kind, etc. and $n$ identical objects of $n^{th}$ kind. There are in total $n^2$ objects.
The number of their arrangements is $\frac{\left(n^2\right)!}{(n!{)}^n}$ which is an integer.
$\left(n^2\right)!$ is divisible by $(n!{)}^n$
Hence, option D is correct.