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.

  • 2
    $\begingroup$ 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 $1=1,\dots,m$. $\endgroup$
    – saulspatz
    Jul 20, 2019 at 3:31
  • $\begingroup$ @saulspatz Sorry I dont know what is combinatorial proof. $\endgroup$
    – DVdivi
    Jul 20, 2019 at 3:35
  • $\begingroup$ @Thomas Andrews: Thank you. It was a typo, fixed. $\endgroup$
    – DVdivi
    Jul 20, 2019 at 6:41

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .