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:
Correct Answer: 6. (i) $\frac{(52) !}{(13 !)^{4}}$ (ii) $\frac{(52) !}{4 !(13 !)^{4}}$ (iii) $\frac{(52) !}{3 !(17)^{3}}$
Solution:
- 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).
