# 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!

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Multiply both sides by $y!$, then replace $x$ with $r$ and $y$ with $n$ and this question becomes a duplicate of math.stackexchange.com/questions/7757/… – anon Sep 29 '11 at 1:02
I misunderstood your question when you posted it as a comment on my earlier answer - I'd thought you were asking about the sum $\displaystyle\sum_{x=0}^y\frac{1}{x\cdot\binom{y}{x}}$. – Zev Chonoles Sep 29 '11 at 1:04

## 5 Answers

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)

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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 $g_2(t) = \exp(t)$.

$$\sum_{x=0}^{y} x \frac{1}{x!} \frac{1}{(y-x)!} = [t]^y ( g_1(t) g_2(t) ) = [t]^y ( t \exp(2 t) ) = \frac{2^{y-1}}{(y-1)!} = \frac{y 2^{y-1}}{y!}$$

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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.

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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 $2^y$.

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Of course, this doesn't work if you're trying to prove that the expected number of times you get a head is $y/2$ unless you do something else in addition. One possibility is to say that it's the sum of the expected values of the number of heads on each trial, each of which is $1/2$. – Michael Hardy Sep 29 '11 at 1:52

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 $m$.

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