Permutations and Combinations 4 Question 6

6. Using permutation or otherwise, prove that $\frac{n^{2} !}{(n !)^{n}}$ is an integer, where $n$ is a positive integer.

(2004, 2M)

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Answer:

N\A

Solution:

  1. Here, $n^{2}$ objects are distributed in $n$ groups, each group containing $n$ identical objects.

$\therefore$ Number of arrangements

$ \begin{aligned} & ={ }^{n^{2}} C _n \cdot{ }^{n^{2}-n} C _n \cdot{ }^{n^{2}-2 n} C _n \cdot{ }^{n^{2}-3 n} C _n \cdot{ }^{n^{2}-2 n} C _n \ldots{ }^{n} C _n \\ & =\frac{\left(n^{2}\right) !}{n !\left(n^{2}-n\right) !} \cdot \frac{\left(n^{2}-n\right) !}{n !\left(n^{2}-2 n\right) !} \cdots \frac{n !}{n ! \cdot 1}=\frac{\left(n^{2}\right) !}{(n !)^{n}} \end{aligned} $

$\Rightarrow$ Integer (as number of arrangements has to be integer).



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