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

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SUMMARY

The discussion focuses on proving the mathematical expression $\displaystyle \frac{\left(n^2\right)!}{\left(n!\right)^n}$ for natural numbers $n$. Participants explore various approaches, including permutations and combinatorial arguments, to establish the validity of this expression. The conversation highlights the significance of factorials in combinatorial mathematics and the implications of this proof for understanding permutations in larger sets. Kali's contribution emphasizes the need for clarity in mathematical proofs and the importance of rigorous argumentation.

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

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 $(n!)^n$
 
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Thanks Kali.

My Solution:

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.
 
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