Simplify $\sum\limits_{a_1=0}^{p_1}\sum\limits_{a_2=0}^{p_2}\sum\limits_{a_3=0}^{p_3}...\sum\limits_{a_n=0}^{p_n}\frac{p!}{a_1!a_2!a_3!.... a_n!}$ I am currently doodling around with some mathematics and stumbled across the following expression:
$$\sum\limits_{a_1=0}^{p_1}\sum\limits_{a_2=0}^{p_2}\sum\limits_{a_3=0}^{p_3}\cdots\sum\limits_{a_n=0}^{p_n}\frac{p!}{a_1!a_2!a_3!\cdots a_n!}$$
where $p=p_1+p_2+p_3+\cdots+p_n$ and $p_i \in \mathbb{N}$
This is a tedious expression to calculate and thus I was wondering whether it is possible to simplify it?
 A: As written, the summand doesn't depend on the variables being summed over, so it is simply equal to
$$p_1p_2 \cdots p_n \cdot \dfrac{p!}{p_1!p_2! \cdots p_n!}$$
which, in turn, is equal to
$$\frac{p!}{(p_1-1)!(p_2-1)! \cdots (p_n-1)!}$$
If, instead, you intended to sum over all $(p_1, p_2, \dots, p_n)$ such that $p_1+p_2+\cdots+p_n=p$, then the answer is quite simply $n^p$.
The reason for this is that $\frac{p!}{p_1!p_2!\cdots p_n!}$ is a multinomial coefficient, which counts the number of partitions of a set $A$ of size $p$ into $n$ sets $(A_1, A_2, \dots, A_n)$, where $|A_i|=p_i$ for each $1 \le i \le n$. Summing over all the possible sizes of such sets, i.e. all $(p_1, \dots, p_n)$ such that $p_1+p_2+\cdots+p_n=p$, the sum thus counts the number of partitions of a set of size $p$.
A partition of $A$ into $n$ sets is equivalent to a function $f : A \to [n]$, where $A_i = f^{-1}(\{i\})$ is the set of elements of $A$ mapped to $i$ by $f$. Thus the sum is equal to the number of functions $A \to [n]$ when $|A|=p$, which is precisely $n^p$.
