combinatorial arguments (2) (1 Viewer)

juantheron · Oct 11, 2014

$$Using\; Combinatorial \; argument \; How\; can\; we\; prove \; $ \displaystyle \frac{n^2!}{(n!)^2}$\; is \; an$

$Integer \; Quantity\;,where \; $n\in \mathbb{N}$

Carrotsticks · Oct 11, 2014

$\\ $Consider a sequence of digits from $ 1 $ to $ n $ where each of these digits occur $ n $ times. Suppose we wish to find the number of possible permutations. \\ \\ There are a total of $ n^2 $ digits and due to repetition, our answer is divided by $ (n!)^n $ so we have $ \frac{(n^2)!}{(n!)^n} $ being an integer. If we multiply this by $ (n!)^{n-2} $ then our result must of course be an integer.$$

combinatorial arguments (2) (1 Viewer)

juantheron

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Carrotsticks

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