I'm trying to get the following sum in terms of n, where the brackets represent nCr of combination theory.
$$ S = \sum_{x=0}^n\sum_{y=0}^{n-x}\sum_{z=0}^{n-x-y} \begin{pmatrix} n \\ x \\ \end{pmatrix} \begin{pmatrix} n-x \\ y \\ \end{pmatrix} \begin{pmatrix} n-x-y \\ z \\ \end{pmatrix} $$
By simple combination logic, I'm expecting S to be 3^n. Any help is appreciated.
JMoravitz is correct in his comment.
You can rewrite the summation as:$$\sum_{x+y+z+u=n}\frac{n!}{x!y!z!u!}$$where $x,y,z,u$ are nonnegative integers.
In this we can recognize the number of ways the set $\{1,\dots,n\}$ can be split up in $4$ covering disjoint and distinguishable subsets.
Approaching this differently we can say that for each of the elements of $\{1,\dots,n\}$ there are $4$ options, leading to $4^n$ possibilities.