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Using permutation or otherwise, prove that (n2)!(n!)n\frac { { (n }^{ 2 })! }{ { (n!) }^{ n } } is an integer without using principle of mathematical induction.

#Algebra #Combinatorics #NumberTheory

Note by Vighnesh Raut
6 years, 12 months ago

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Comments

Let us find the number of ways dividing n2n^2 distinct items into nn groups of nn items each. Let the number be NN.

Case 1: If the order matters or the groups are distinguishable in the sense that for example, out a group of 44 items, say, (A,B,C,D)(A, B, C, D) the groups ((AB),(CD))((AB) , (CD)) and ((CD),(AB))((CD) , (AB)) are considered to be different, then

N=(n2)!(n!)(n!)…(n times )…(n!)=(n2)!(n!)n \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 NN 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!n! as that is the number of ways of arranging the nn groups which does not matter here. Hence,

N=(n2)!(n!)n×1(n!)=(n2)!(n!)n+1 \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 - 6 years, 11 months ago
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