factorial!!! (2 Viewers)

Kathy358 · Jan 29, 2018

Proved that, if n and p are positive integers, (np)! is a multiple of (p!)ⁿn!
Please help me!. Thanks a lot!

InteGrand · Jan 29, 2018

Kathy358 said:
Proved that, if n and p are positive integers, (np)! is a multiple of (p!)ⁿn!
Please help me!. Thanks a lot!

$\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.)$

factorial!!! (2 Viewers)

Kathy358

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InteGrand

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