# Prove that $~~~~~a_1!a_2!\cdots a_m! \mid \left(a_1+a_2+...+a_m\right)! ~~~~~~~~~\forall ~~~a_1,a_2,...,a_m\in N$

Show that $$a_1!a_2!\cdots a_m! \mid \left(a_1+a_2+...+a_m\right)! \forall a_1,a_2,...,a_m\in N$$

Case 1 $$m=2$$. We need to prove $$a_1!a_2!\mid (a_1+a_2)!$$

1. $$a_1+a_2=1$$ and $$a_1+a_2=2$$. It is obvious

2. $$a+b=n (n\in \text{N and } a+b\le n).$$

We will prove in $$a+b=n+1$$

By assuming of the induction $$(a_1!(a_2-1)!)\mid(a_1+a_2-1)!$$

And $$(a_2!(a_1-1)!)\mid(a_1+a_2-1)!$$

$$\rightarrow (a_1!a_2!)\mid (a_1+a_2)(a_1+a_2-1)!$$

Or $$\rightarrow (a_1!a_2!)\mid (a_1+a_2)!$$

Case 2 $$m=k$$

We will prove in $$m=k+1$$

I am stuck here. Can I prove it as with the case $$m = 2?$$, help me.

• If you are allowed to use a combinatorial proof, you can argue that ${(a_1+\cdots+a_m)!\over a_1!\dots a_m!}$ is the number of ways to distribute $a_1+\cdots+a_m$ distinct objects into $m$ labeled bins, with $a_i$ objects in bin $i$ for $1=1,\dots,m$. Jul 20, 2019 at 3:31
• @saulspatz Sorry I dont know what is combinatorial proof. Jul 20, 2019 at 3:35
• @Thomas Andrews: Thank you. It was a typo, fixed. Jul 20, 2019 at 6:41

By the $$m=2$$ case: $$a_1!(a_2+\ldots+a_{n+1})!\,\vert\, (a_1+a_2+\ldots+a_{n+1})!,$$ and by the inductive hypothesis for $$m=n$$, $$a_2!\ldots a_{n+1}!\vert (a_2+\ldots+a_{n+1})!.$$ Putting these together gives the $$n+1$$ case.