Let $a,b,n$ be natural numbers such that $a!b! \mid n!$. I am looking for a (somehow best) upper bound of $a+b$ in terms of $n$ (for large values on $n$). For example it is clear to see that we must have $a+b \leq 2n$. But unfortunately I am looking for much smaller bound ! Any idea would be helpful. 



How $a+b$ can grow when $a!b! \mid n!$
This is a small variation of Seva's argument yielding a slightly more explicit result. 


Assume that $a+b>n$ with $a!b!$ dividing $n!$. Then
$$ \prod\_{n-a<m\leq b}m=\frac{b!}{(n-a)!} \ \ \text{divides}\ \ \frac{n!}{(n-a)!a!}= \binom{n}{a}. $$
Looking at the exponent of $2$ in the prime decomposition of the two sides, we infer (e.g. by [Kummer's theorem](http://planetmath.org/KummersTheorem))
$$ \frac{b-(n-a)-1}{2} \leq \log\_2(n),$$
whence
$$ a+b-n \leq 1+2\log\_2(n). $$
Of course the same result is trivial when $a+b\leq n$.


