Prove that $(kn)!$ is divisible by $(k!)^n$

Question

Suppose $k,n$ are integers $\ge1$. Show that $(kn)!$ is divisible by $(k!)^n$

I have simplified the problem and now, I need to prove that the product of any $k$ consecutive integers is divisible by $k!$. However I am stuck there.

Answer 1

$k!|(1.2.3.\dots k)$

$k!|(k+1).(k+2)\dots(2k )$

and so on till,

$k!|((n-1)k+1)((n-1)k+2)\dots (nk)$

so we have the result.

We know that the number of ways of choosing $k$ objects out of a collection of $n$($n\ge k$) identical objects is $n \choose k$(it an easily be proved) which from the definition is an integer.

But $\displaystyle {n\choose k}=\frac{(n-k+1).(n-k+2)\dots (n)}{k!}$

So we can conclude from this that every product of k consecutive integers is divisible by $k!$

Answer 2

You may not have done group theory yet, but here is how it is possible to use Lagrange's Theorem to prove the result you want (and a slightly stronger one). Lagrange's Theorem says that the order of a subgroup $H$ of a finite group $G$ divides the order of $G$- it's one of the most basic and important theorems in finite group theory.

An important finite group is the symmetric group $S_{m},$ the group of all possible permutations of $m$ objects, with group operation composition. The symmetric group $S_{nk}$ has a subgroup which is the direct product of $n$ copies of $S_{k},$ where the $i$-th factor is the group of all permutations of $\{1+(i-1)n, \ldots, k +(i-1)n \}$ (fixing all the remaining points). This subgroup (ie, the whole direct product) clearly has order $(k!)^{n},$ so Lagrange's Theorem gives what you want.

In fact, we can do a little bit better. There is a subgroup $S_{k} \wr S_{n}$ of $S_{kn}$, which has that direct product as its "base group", and then we permute these $n$ blocks of size $k$ around as $S_{n}$ does (treating the blocks of size $k$ as "points"). This group is still a subgroup of $S_{kn},$ and is the largest subgroup of $S_{kn}$ which preserves the chosen decomposition into $n$ blocks of size $k.$ It has order $n!(k!)^{n}.$ Hence it is in fact the case that $n!(k!)^{n}$ divides $(nk)!$.

Answer 3

One answer has been given already. Here is another: $$ \frac{(k\cdot n)!}{(k!)^n}=\binom{k\cdot n}{k,k,\ldots,k} $$ (with $n$ occurrences of $k$ on the right) is a multinomial coefficient, known to be an integer.

More generally, $$ \binom{k_1+k_2+\cdots+k_n}{k_1,k_2,\ldots,k_n} =\frac{(k_1+k_2+\cdots+k_n)!}{k_1!\cdot k_2!\cdots k_n!} $$ is the number of ways to select $n$ subsets $S_1$, …, $S_n$ out of a set of $k_1+k_2+\cdots+k_n$ members, with $S_j$ having cardinality $k_j$.

You can see this by direct combinatorial reasoning: You can specify such a selection just by ordering the big set, then taking $S_1$ to be the first $k_1$ members in the ordering, and so on. There are $(k_1+k_2+\cdots+k_n)!$ to do this, but then you have overestimated by a factor $k_1!\cdot k_2!\cdots k_n!$, since permuting the members of any $S_j$ does not change the selection.