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!

For a combinatorial proof of the non-obvious step in Zev’s argument, $$\sum_{x=0}^y x\binom{y}{x} = y2^{y-1},$$ suppose that you have $y$ children, and you want to choose a team (of any size) from the group. However, a team is required to have a captain, and two teams are counted differently if they have different captains, even if they have exactly the same members.
For any $x$ there are $\dbinom{y}{x}$ ways to choose $x$ children to form a team, and there are then $x$ ways to choose the captain of the team, so $x\dbinom{y}{x}$ is the number of ways of choosing a ‘captained’ team of $x$ players. Thus, the sum on the left-hand side of the equation gives the total number of possible ‘captained’ teams.
On the other hand, we could first choose a captain and then choose the rest of the team. There are $y$ ways to choose a captain. Once the captain has been chosen, there are $2^{y-1}$ subsets of the remaining $y-1$ children that could form the rest of the team, so there are $y2^{y-1}$ ways to form a ‘captained’ team.
