Summation of reciprocal of Product of Factorials.

Question

How can this summation be evaluated: $${∑ {1\over {a_1!a_2!....a_m!}}}$$

Where $$a_1+a_2+.....+a_m=n$$
Also $a_i !=n $ and $m<n$.

Answer 1

$\sum\frac{n!}{\prod_ia_i!}=\sum\frac{n!}{\prod_ia_i!}\prod_ix_i^{a_i}$ with $x_i=1$;

hence $\sum\frac{n!}{\prod_ia_i!}=(\sum_i x_i)^{a_1+\ldots+a_m}=m^n$.

Finally, the result is $m^n/n!$.