Number Theory Question Possibly related to combinatorics too.

[iironiic]

Homework Statement

Prove that $a! b! | (a+b)!$.

Homework Equations

Probably some Number Theory Theorem I can't think of at the moment.

The Attempt at a Solution

Without loss of generality, let $a < b$.
Therefore $b! | \Pi _{k=1}^b a+k$. Since $(a+1) ... (a+b)$ are $b$ consecutive terms, $b|\Pi _{k=1}^b a+k$. The problem I am running into is that you don't necessarily get $b-1$ consecutive terms in the next step of the recursive reasoning. Any suggestions? Thanks!

This question reminds me of questions like, how many possible combinations can you rearrange the letters in "MISSISSIPPI"? This might be a combinatorics question too but I'm not entirely sure.

 

[kru_]

Consider that (a+b)! = (1)(2)(3)...(a)(a+1)(a+2)...(a+b).

And a! = (1)(2)(3)...(a).

Now it should be clear that (a+b)!/a! = (a+1)(a+2)...(a+b). Then your assertion that b! divides that should be sufficient.