Integer Quantity Prove: Permutations/Otherwise n^2! / (n!)^n

[juantheron]

Prove by permutations or otherwise $\displaystyle \frac{\left(n^2\right)!}{\left(n!\right)^n}$, where $n\in \mathbb{N}$

 

[kaliprasad]

Prove by permutations or otherwise $\displaystyle \frac{\left(n^2\right)!}{\left(n!\right)^n}$, where $n\in \mathbb{N}$

Spoiler

we know that product of n consecutive numbers is divisible by n!

so n! is dvisible by n!
$\frac{\left(2n\right)!}{\left(n!\right)}$ is divisible by n!
$\frac{\left(3n\right)!}{(2n)!}$ is divisible by n!
...
$\frac{\left(n\right)^2!}{(n(n-1))!}$ is divisible by n!

hence the product of 1st term in each of lines that is $(n^2)!$ is divisible product of 2nd term in each of lines that is $(n!)^n$

 

[juantheron]

Thanks Kali.

My Solution:

Spoiler

Distribution of $(mn)$ different objects things into $m$ group each contain $n$ objects $\displaystyle = \frac{(mn)!}{(n!)^m}$

Where order of the group is important.

Now put $m=n\;,$ we get...

Distribution of $(n^2)$ different objects things into $n$ group each contain $n$ objects $\displaystyle = \frac{(n^2)!}{(n!)^n}$

Where order of the group is important.