1. ## factorial!!!

Proved that, if n and p are positive integers, (np)! is a multiple of (p!)nn!

2. ## Re: factorial!!!

Originally Posted by Kathy358
Proved that, if n and p are positive integers, (np)! is a multiple of (p!)nn!
$\noindent \textbf{Hints.} Let p = m (sorry, I started writing with m instead of p, and forgot it was p until I'd written up a fair amount). Note that we can break up (mn)! into n chunks of m consecutive integers:$

$(mn)! = (\color{red}{1}\color{black}m\times (m-1)\times \cdots \times 1)\cdot (\color{red}{2}\color{black}m \times (2m-1) \times \cdots \times (m+1)) \cdot (\color{red}{3}\color{black}m \times (3m-1)\times \cdots \times (2m+1))\cdot (\color{red}{4}\color{black}m \times (4m-1)\times \cdots \times (3m+1)) \cdot \ldots \cdot (\color{red}{n}\color{black}m\times (nm-1)\times \cdots \times (nm-m+1)).$

$\noindent The \color{red}{red} \color{black} numbers above show us a factor of n! in (mn)!, so if we can show that the remaining numbers give us a total of n factors of m!, we will be done. So looking at each of the n chunks of m consecutive integers written above, we see that it suffices to show that each of these chunks when the red number part is excluded is divisible by m!. That is, it suffices to show the following fact:$

$\noindent \color{blue}{For every positive integer k, the product m(km-1)(km-2)\ldots (km-m+1) is divisible by m!.}$

$\noindent Some steps to show the above fact:$

$\noindent \bullet Show that any product of m consecutive integers is divisible by m! (for all positive integers m). (A hint to show this: write such a product as something like a(a-1)(a-2)\ldots (a-m+1) and write this in terms of a binomial coefficient.)$

$\bullet Observe that the above bullet point's result implies that the numbers$

$n_{1}:= km(km-1)(km-2)\ldots (km-m+1) \quad \text{and}\quad n_{2} := (km-1)(km-2)\ldots (km-m+1)(km-m)$

$\noindent are divisible by m!, and hence so is their difference. (Or more simply, the first bullet point implies that (km-1)(km-2)\ldots (km-m+1) is divisble by (m-1)!, since this is a product of (m-1) consecutive integers; this implies the \color{blue}{blue} \color{black} fact.)$

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