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# PMI not allowed...

Using permutation or otherwise, prove that $$\frac { { (n }^{ 2 })! }{ { (n!) }^{ n } }$$ is an integer without using principle of mathematical induction.

Note by Vighnesh Raut
2 years, 4 months ago

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Let us find the number of ways dividing $$n^2$$ distinct items into $$n$$ groups of $$n$$ items each. Let the number be $$N$$.

Case 1: If the order matters or the groups are distinguishable in the sense that for example, out a group of $$4$$ items, say, $$(A, B, C, D)$$ the groups $$((AB) , (CD))$$ and $$((CD) , (AB))$$ are considered to be different, then

$\displaystyle N = \frac{(n^2)!}{(n!)(n!) \dots ( n \text{ times } ) \dots (n!)} = \frac{(n^2)!}{(n!)^n}$

Since the number of ways are always an integer, therefore $$N$$ is an integer.

Case 2: If the order does not matter, then the above result can be modified as follows: The number of ways have to be reduced by a factor of $$n!$$ as that is the number of ways of arranging the $$n$$ groups which does not matter here. Hence,

$\displaystyle N = \frac{(n^2)!}{(n!)^n} \times \frac{1}{(n!)} = \frac{(n^2)!}{(n!)^{n+1}}$

Now if the above is an integer then definitely the expression is. · 2 years, 4 months ago