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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. Sudeep Salgia · 2 years, 4 months ago

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