I think there are $\binom{b+ c - 1}{c-1}$ ways to distribute $b$ balls in $c$ containers. (Please correct me if that's a mistake.) How does this change if I am not allowed to place more than $n$ balls in any given container?
If we call this number $N(b,c,n)$ then I can come up with
$N(b,c,n) = \sum_{k=0}^n N(b-k, c-1, n)$
with the base cases
$N(b,c,n) = \binom{b+ c - 1}{c-1}$
for $n\geq b$, and
$N(b,c,n) = 0$
for $n.
Is there some better way of writing this than that recursion relation?