Inclusion/Exclusion Principle for indistinguishable balls into two different types of boxes

Question

The problem asks me how to distribute n balls into h boxes where each must get at least 1 n and t boxes where each can be left empty

What I have, and where I seem to have stalled is how to incorporate the t boxes, the formula I have right now for h boxes is (n+h-1) choose h

Which I believe is the correct formula for making sure that each h gets at least 1 n, but I have no idea how to incorporate t boxes into this, if anyone could point me in the right direction it would be much appreciated.

Answer 1

It is not quite clear what is intended, so we rephrase. There are $h+t$ (distinguishable) boxes $A_1$ to $A_h$ and $B_1$ to $B_t$. The $A_i$ must each have at least one ball, but some or all of the $B_j$ can be left empty.

Put a ball into each $A_i$. That leaves $n-h$ balls, to be distributed among $h+t$ boxes with no restriction. That is a standard problem.

Answer 2

Hint: let there be $k$ balls in the $h$ boxes, where clearly $k \ge h$. How many ways to distribute them? You have that above. Now for each of those you distribute $n-k$ balls into $t$ boxes. Sum over $k$ from $h$ to $n$.