summation of x * (y choose x) binomial coefficients What does this summation simplify to?
$$ \sum_{x=0}^{y} \frac{x}{x!(y-x)!} $$
I was able to realize that it is equivalent to the summation of $x\dbinom{y}{x}$ if you divide and multiply by $y!$, but I am unsure of how to further simplify.
Thanks for the help!

Using generating function technique as in answer to your other question:
Using $g_1(t) = t \exp(t) =  \sum_{x=0}^\infty t^{x+1} \frac{1}{x!} = \sum_{x=0}^\infty t^{x+1} \frac{x+1}{(x+1)!} = \sum_{x=-1}^\infty t^{x+1} \frac{x+1}{(x+1)!} = \sum_{x=0}^\infty t^{x} \frac{x}{x!}$ and $g_2(t) = \exp(t)$.
$$
  \sum_{x=0}^{y} x \frac{1}{x!} \frac{1}{(y-x)!} = [t]^y ( g_1(t) g_2(t) ) = [t]^y ( t \exp(2 t) ) = \frac{2^{y-1}}{(y-1)!} = \frac{y 2^{y-1}}{y!} 
$$
