Today I was thinking about this problem:
I have $n\geq2$ boxes and $m\geq1$ balls. I know that the $i-$th box can contain at most $l_i\geq1$ balls and moreover I know that $l_1+\dots+l_n=2m.$
I want to compute the possible different combination in which I can fill the boxes.
Is it possible? Can we at least find a rasonable lower bound? (for example when $m$ is big).
This is just an approach, not a complete solution.. You need the number of solutions to $$ \sum_{i=1}^n x_i= m\quad \textrm{ with } x_i \in [0,\ell_i] \cap \mathbb{Z}\quad \forall i\in [n]$$ This is the same as the coefficient of $t^m$, in the expansion of \begin{align} \prod_{i=1}^n \left(1+ t + t^2 + \ldots +t^{\ell_i}\right) &=\prod_{i=1}^n \frac{1-t^{1+\ell_i}}{1-t}\\ &=\left[ \sum_{k=0}^\infty \binom{n+k-1}{k} t^k\right] \prod_{i=1}^n (1-t^{1+\ell_i})\\ &=1 + nt+ \left(\cdots\right)t^2 + \cdots +\left[\binom{n+m-1}{m} + \ldots \right]t^m + \cdots \end{align} If you can calculate the rest of the terms in the square bracket above, you are done.