SATHEE: Permutations and Combinations 4 Question 6

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.

$∴$ Number of arrangements

$\begin{aligned} =^{n^{2}} C_{n} \cdot^{n^{2} - n} C_{n} \cdot^{n^{2} - 2 n} C_{n} \cdot^{n^{2} - 3 n} C_{n} \cdot^{n^{2} - 2 n} C_{n} \dots^{n} C_{n} \\ = \frac{(n^{2})!}{n! (n^{2} - n)!} \cdot \frac{(n^{2} - n)!}{n! (n^{2} - 2 n)!} \dots \frac{n!}{n! \cdot 1} = \frac{(n^{2})!}{(n!)^{n}} \end{aligned}$

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

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Permutations and Combinations 4 Question 6

6. Using permutation or otherwise, prove that n2!(n!)n is an integer, where n is a positive integer.

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