Constrained Sum of Factorials

Question

Consider the sum $$S=\sum_{\substack{r_i>0,\\1\le i\le m\\ r_1+...+r_m=n}}\frac{1}{r_1!...r_m!},$$ where $m,n$ are fixed, positive integers, and the $r_i$ are integers. If there were no condition that $r_1+...+r_m=n$, then this could be decomposed as $$\sum_{\substack{r_i>0,\\1\le i\le m}}\frac{1}{r_1!...r_m!}=\prod_{i=1}^m\sum_{r_i>0}\frac{1}{r_i!}=(e-1)^m$$ But with the condition $r_1+...+r_m=n$, I am unsure how to sum this. I suspect one could, for example, use $$\sum_{n=1}^k \frac{1}{n!}=e\frac{\Gamma(k+1,k)}{k!}-1,$$ where $\Gamma$ is the incomplete gamma function, but I am unsure how to proceed. How would one go about getting an expression for this?

Answer 1

Perform a change of variables $s_i=r_i-1$ and apply the multinomial theorem: \begin{align} S&=\sum_{\substack{r_i>0,\\1\le i\le m\\ r_1+\dots+r_m=n}}\frac{1}{r_1!\dots r_m!} \\ &=\sum_{\substack{s_i\ge 0,\\1\le i\le m\\ s_1+\dots+s_m=n-m}}\frac{1}{s_1!\dots s_m!} \\ &=\frac{1}{(n-m)!}\sum_{\substack{s_i\ge 0,\\1\le i\le m\\ s_1+\dots+s_m=n-m}}\frac{(n-m)!}{s_1!\dots s_m!} \\ &=\frac{m^{n-m}}{(n-m)!} \end{align}