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# Factorial

Prove that (n!)! Is divisible by (n!)^(n-1)!

Note by Parv Mor
3 years, 6 months ago

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## Comments

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The number of ways to arrange $$k = k_1 + k_2 + \ldots+ k_m$$ items where $$k_1$$ are of one type, $$k_2$$ of another type etc is $$\dfrac{k!}{k_1!k_2! \dots k_m!}$$. Now this is obviously an integer as it denotes the number of ways to arrange objects.

Putting $$k = n!, k_1 = k_2 = \dots = k_{(n-1)!} = n$$, we see that it satisfies $$k = k_1 + k_2 + \ldots+ k_m$$ and number of arrangements is $$\dfrac{(n!)!}{n!n! \dots n!} = \dfrac{(n!)!}{n!^{(n-1)!}}$$. Since this is an integer, this implies $$(n!)^{(n-1)!}$$ divides $$(n!)!$$

EDIT: Changed all the $$(n-1)$$ to $$(n-1)!$$

- 3 years, 6 months ago

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How is k{1} +k{2} +..... k_{n-1}= k ,i.e, n as it must be equal to n(n-1). Also question for (n!)! / n!^{(n-1)!} not for (n!)! / n!^{(n-1)}

- 3 years, 6 months ago

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Yeah. Sorry. I'd put $$(n-1)$$ instead of $$(n-1)!$$. Sorry again. -.-

- 3 years, 6 months ago

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