# Number Theory Question! Possibly related to combinatorics too.

1. Mar 26, 2012

### iironiic

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Mar 26, 2012

### 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.