Let $d,m, r$ be positive integers, and define $$ S = \left\{ (i_1, i_2, \dots, i_m) \in {\bf Z}_{+}^{m} \left | \sum_j i_j = d; \& \forall j, i_j \leq r \right. \right\}; $$ Here ${\bf Z}_+$ denotes the set of nonnegative integers (that is, including zero). So we are taking all (ordered) integer partitions of $d$ with $m$ constituents (permitting zero as a constituent), with the constituents bounded above by r.

I would like a good upper bound (in terms of $m$, $d$, and $r$) for $$ \sum_{(i_j) \in S} \frac 1{i_1! \cdot i_2! \cdot \dots \cdot i_m!}. $$ When there is no $r$ condition (e.g., if $r \geq d$), then it looks like the sum is exactly $m^d/d!$, arising from representing $m^d$ as a combination of descending products. And of course, if $mr < d$, then there are no terms.

However, I am most interested in the case when $d$ is less than $mr$ (special cases: if $d = mr-1, mr-2, mr-3, mr-4, $, etc.; exact results are easily obtained). But I would like a general result or a reference for this type of problem (with $d$ much less than $mr$); this must have been considered before.