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!
 A: This problem is not too bad: we have
$$\sum_{x=0}^y\frac{x}{x!(y-x)!}=\frac{1}{y!}\sum_{x=0}^y\frac{x\cdot y!}{x!(y-x)!}=\frac{1}{y!}\sum_{x=0}^yx\cdot \binom{y}{x}=\frac{y2^{y-1}}{y!}=\frac{2^{y-1}}{(y-1)!}$$
(see the second equation here)
A: 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 $g_2(t) = \exp(t)$.
$$
  \sum_{x=0}^{y} x \frac{1}{x!} \frac{1}{(y-x)!} = [t]^y ( g_1(t) g_2(t) ) = [t]^y ( t \exp(2 t) ) = \frac{2^{y-1}}{(y-1)!} = \frac{y 2^{y-1}}{y!} 
$$
A: 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.
A: If you toss a coin $y$ times, the probability of getting a head exactly $x$ times is $\dbinom yx 2^{-y}$.  And the expected number of times you get a head is obviously $y/2$.  But the expected number of times you get a head is also
$$
\sum_{x=0}^y x\binom yx 2^{-y}.
$$
So
$$
\sum_{x=0}^y x\binom yx 2^{-y} = \frac y2.
$$
Now multiply both sides by $2^y$.
A: For a non-combinatorial evaluation of $\sum_{x=0}^y x\binom yx $, consider
$\sum_{x=0}^y z^x \binom yx $. This is just $(1+z)^y$. Differentiate this and set $z = 1$.
BTW, my mind complains at having $x$ and $y$ where I am used to seeing $n$ and $m$.
