Prove that $(an+bm)!$ is divisible by $m!.n!$

Question

I do not know how to go about proving this question. The answers I have received so far seem to have ignored that the factorial of $an+bm$ must be divisible by the product of the factorials of $m$ and $n$. It is given that $a,b,n,m \in\ \mathbb{Z} ^+$

Answer 1

$$\frac{\left(an+bm\right)!}{m!\cdot n!}=\binom{an+bm}{m}\cdot\binom{an+\left(b-1\right) m}{n}\cdot\left(\left(a-1\right)n+\left(b-1\right)m\right)!$$ is the product of three integers, and therefore an integer.

Here, $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ represents the number of ways to choose $k$ objects from a set of $n$. By this definition, this number is always an integer.

Answer 2

Not sure when $a,b,m,n\in\mathbb{Z}$ but yes, we can prove it when $a,b,m,n\in\mathbb{Z}^+$. Note that $m\choose{n}$ is an integer and hence, $m!n!|(m+n)!$. Now, for any $a,b\in\mathbb{Z}^+$, $(m+n)!|(am+bn)!$ because $am+bn\geq m+n$. Thus, $m!n!|(am+bn)!$. Indeed, if you allow $a,b,m,n\in\mathbb{Z}$, then, as pointed out by @Sabhrant this is not true.

Answer 3

$(an+bm)! = (\prod_{j=1}^n j)(\prod_{k=n+1}^{n+m} k)(\prod_{i=n+m+1}^{an+bm} i)$

Now $n! | (\prod_{j=1}^n j)$ and $m!|(\prod_{k=n+1}^{n+m} k)$.

So $m!n!| (\prod_{j=1}^n j)(\prod_{k=n+1}^{n+m} k)$ and so $m!n!| (\prod_{j=1}^n j)(\prod_{k=n+1}^{n+m} k)(\prod_{i=n+m+1}^{an+bm} i)$

.... that's all there is to it.....

Why does $n!|(\prod_{j=1}^n j)$? Because $(\prod_{j=1}^n j)=1*.....*n = n!$.

Why does $m!|(\prod_{k=n+1}^{n+m} k)$. Because $(\prod_{k=n+1}^{n+m} k)= (n+1)* .... *(n+m)$ is the product of $m$ consecutive (positive) integers, and the product of $m$ consecutive products is divisible by $m$!

Why?

Lemma: For any set of $m$ consecutive positive integers, the $\prod_{i=1}^m (b+i)$ is divisible by $m!$.

Why? $\frac {\prod_{i=1}^k (b+i)}{k!} = \frac {(b+1)...(b+k)}{k!} = \frac {(b+k)!}{b!k!} = {b+k \choose k}$ which is literally the number of ways to pick $k$ objects from a collection of $b+k$ objects. And the standard derivation of that, which I assume you have seen, proves that. Well, the number of ways to do that must be an integer.

Answer 4

Note that for $a,b,m,n\in \mathbb Z_+$: $$\frac{(an+bm)!}{m!n!}=\frac{(an+mb)\cdot (an+mb-1)\cdots(n+m+1)(n+m)!}{m!n!}=\\ (an+mb)(an+mb-1)\cdots(n+m+1){n+m\choose m},$$ which is a positive integer as the factors are such.