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I try $n=\overbrace{m+m+m+\cdots +m}^{\text{k times}}$ then multinomial coefficient is $\underbrace{\dfrac{n!}{m!m!\cdots m!}}_{\text{k times}} = {n \choose m,m,\cdots,m}$ but now I'm stuck.

Fawkes4494d3
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Alberto Chavez
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  • Do you mean the multinomial coefficient maybe? Binomial means there will be two terms in the sum as $n$, i.e. $n=a+b$ is a binomial form for $n$, so $\dfrac{n!}{a!(n-a)!}=\dfrac{n!}{a!b!}$ is a binomial coefficient. The one you want to prove is an integer for $k>2$, is a multinomial coefficient. Refer to https://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients – Fawkes4494d3 Jul 01 '20 at 03:02
  • I've edited your post a bit for better reading, but do refer to the link above and let know if it helps. – Fawkes4494d3 Jul 01 '20 at 03:08
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    HINT: The multinomial coefficient $$\binom{n}{\underbrace{m,\ldots,m}_k}=\frac{n!}{m!^k}$$ is an integer because it gives the number of ways to ... what? – Brian M. Scott Jul 01 '20 at 03:11
  • @Fawkes4494d3 oh yes multinomial coefficient i want to mean – Alberto Chavez Jul 01 '20 at 03:13
  • @BrianM.Scott Permutations i think so – Alberto Chavez Jul 01 '20 at 03:14
  • @AlbertoChavez: Not quite; you might take a look here. – Brian M. Scott Jul 01 '20 at 03:18

2 Answers2

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One approach for people who might be unfamiliar with the multinomial coefficient is to use induction on k.

When $k=1$, $n=m$ so $\frac{n!}{(m!)^k} = 1$ which is clearly an integer.

Suppose that for $k = i$, for all $n,m$ s.t. $n = km$, $\frac{n!}{(m!)^k}$ is an integer.

Note that as $n=km$ and $k$ is a positive integer, $m \leq n$.

Now, suppose $n = (i+1)m$. Then $n = im + m$. So $n-m = im$.

\begin{align*}\frac{n!}{(m!)^{(i+1)}} & = \frac{(n..(n-m+1) \times (n-m)!)}{m!^i \times m!} \\ & = \frac{n..(n-m+1)}{m!} \times \frac{n-m}{m!^i}\end{align*}

By the inductive hypothesis, we see that the right side of this product is an integer. It remains to prove that the left side is also an integer.

Note that the left side is actually just the binomial coefficient ($n$ choose $m$). So it is clearly an integer.

There are also arithmetic proofs that the left side is an integer, but they're very involved. Check the discussion here if interested:

The product of $n$ consecutive integers is divisible by $n$ factorial

Drathora
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Hint:

Since ${m \choose m},\,{2m \choose m},\,\ldots\,,{(k-1)m \choose m}\;\text{and}\;{km\choose m}$ are all integers, so is their product $${m \choose m}{2m \choose m}\ldots\,{(k-1)m \choose m}{km\choose m}$$

Ángel Mario Gallegos
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