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.



Sums of reciprocals of products of factorials
I'll just make my comment above an answer. The sum in question multiplied by $d!/m^d$ can be interpreted as the probability that when $d$ balls are thrown into $m$ boxes then each box contains no more than $r$ balls (the balls are thrown into boxes at random uniformly and independently).


This problem arose earlier on MathOverflow in connection with [Elkies's coupon collector's earworm question](https://mathoverflow.net/questions/215664/the-coupon-collectors-earworm). From the answer there, I link [the paper by Raab and Steger](http://www14.informatik.tu-muenchen.de/personen/raab/publ/balls.pdf), and Theorem 1 there gives information on the behavior of this probability (or equivalently of your sum). There is a sharp range of $r$ in which the sum transitions from $o(m^d/d!)$ to $(1-o(1))m^d/d!$. 


