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This question already has an answer here:

Prove that $\sum_{r=1}^n r\binom n r^2 = n\binom{2n-1}{n-1}$

I tried: $\sum_{r=1}^n r\binom n r^2 = n\sum_{r=1}^n \binom {n-1}{r-1}\binom n r$ using the identity $\binom n r = \frac n r \binom {n-1}{r-1}$ for $r \ge 1$.

Not sure how to proceed from there. I tried expanding the RHS but can't seem to find a way to cancel out the factorials.

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marked as duplicate by lab bhattacharjee, Empty, Servaes, Mankind, Tom-Tom Sep 24 '15 at 9:39

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So we want to show that $$\sum^n_{r=1}\binom{n-1}{r-1}\binom{n}{r}=\binom{2n-1}{n-1},$$ or $$\sum^n_{r=1}\binom{n-1}{n-r}\binom{n}{r}=\binom{2n-1}{n}.$$ Now, how many ways to choose a committee of $n$ people out of $n-1$ males and $n$ females? Consider the cases where there are $r$ females in the committee.

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  • $\begingroup$ I see what you mean. Here you're giving a combinatorial argument. Is there an algebraic way to complete my unfinished proof above? By the way, I think you made a typo in the fourth line, it should be $\binom {2n-1}{n-1}$ right? $\endgroup$ – Vizuna Sep 24 '15 at 4:59
  • $\begingroup$ No, it's $n$: $\binom{2n-1}{n-1}=\binom{2n-1}{n}$. And I don't really know any algebraic proof. Combinatorics just works so well. $\endgroup$ – Quang Hoang Sep 24 '15 at 5:21
  • $\begingroup$ But the only difference between the 2nd line and the 4th line is changing $\binom{n-1}{r-1}$ to $\binom{n-1}{n-r}$. Since they are equal, how come the RHS changes? $\endgroup$ – Vizuna Sep 24 '15 at 5:27
  • $\begingroup$ I don't understand: The RHS of the two lines also have the same value. That "change" is to serve the combinatoric argument. $\endgroup$ – Quang Hoang Sep 24 '15 at 5:31
  • $\begingroup$ Oh, right. Sorry, I missed that. $\endgroup$ – Vizuna Sep 24 '15 at 5:34
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A shop has $n$ different doughnuts and $n$ different muffins. I want to choose a total of $n$ items, including at least one doughnut, and I would like to have one doughnut to eat right away, and the remaining $n-1$ put in a bag. How many ways can this be done?

The doughnut I eat right away can be chosen in $n$ ways, and for each of these ways there are $\binom{2n-1}{n-1}$ ways to choose the remaining items.

Let us count another way. If we decide to get $r$ doughnuts, that can be done in $\binom{n}{r}$ ways, and then the doughnut we eat right away can be chosen in $r$ ways. The $n-r$ muffins can be chosen in $\binom{n}{n-r}$ ways, that is, in $\binom{n}{r}$ ways. So there are $r\binom{n}{r}\binom{n}{r}$ to carry out the task while choosing $r$ doughnuts. Add up, $r=1$ to $n$.

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  • $\begingroup$ Thanks! Would it be correct to say that your argument is similar to Quang Hoang's argument above where at least 1 female must be in the committee? $\endgroup$ – Vizuna Sep 24 '15 at 5:21
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    $\begingroup$ They are both of a combinatorial character. The one by Quang Hoang does a preliminary transformation of $r\binom{n}{r}$ to $n\binom{n-1}{r-1}$. That makes the "story" simpler than mine, which attempts to preserve the combinatorial meaning of $n\binom{2n-1}{n-1}$ and the terms on the original left-hand side. $\endgroup$ – André Nicolas Sep 24 '15 at 5:30
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Using Vandermonde's Identity, we get $$ \begin{align} \sum_{r=0}^nr\binom{n}{r}^2 &=\sum_{r=0}^nr\binom{n}{r}\binom{n}{n-r}\\ &=n\sum_{r=0}^n\binom{n-1}{r-1}\binom{n}{n-r}\\ &=n\binom{2n-1}{n-1} \end{align} $$

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  • $\begingroup$ But to use Vandermonde's Identity shouldn't it be $n\sum_{r=0}^n\binom{n-1}{r}\binom{n}{n-r}$ rather than $n\sum_{r=0}^n\binom{n-1}{r-1}\binom{n}{n-r}$? $\endgroup$ – Vizuna Sep 24 '15 at 5:15
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    $\begingroup$ @Vizuna: it is just a change of variables: $$n\sum_{r=0}^n\binom{n-1}{r-1}\binom{n}{(n-1)-(r-1)} = n\binom{(n-1)+n}{n-1}$$ The way I think of Vandermonde's Identity is that the upper entry is the sum of the upper entries and the lower entry is the sum of the lower entries, as long as we've summed over all the possible values of the lower entries. $\endgroup$ – robjohn Sep 24 '15 at 10:03

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