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Give a combinatorial proof & proof using generating functions of the following identity:

$$\sum_{k=1}^n k {n\choose k}^2 = n {2n-1 \choose n-1}.$$

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closed as not a real question by Andres Caicedo, t.b., Eric Naslund, Zev Chonoles, Guess who it is. Apr 26 '11 at 12:54

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

Is this homework? –  Phira Apr 23 '11 at 20:56
Do you know what the terms mean? Can you give us some background? Do you know the corresponding proofs of the Vandermonde identity? –  Phira Apr 23 '11 at 20:59
Well, if you don't know whether you know what $C(n,k)$ means or what the summation index of your sum is, or whether you know the corresponding proof of the Vandermonde identity, I don't know why you are asking the question. Do you think that I will explain your question to you? –  Phira Apr 23 '11 at 21:15
As you alluded, it looks like the Vandermonde identity. Look at the proofs (combinatorial and generating function respectively) of that identity and see how you can deal with the $k$. –  Mitch Apr 23 '11 at 21:26
Please do not post questions in the imperative mode. You are not assigning us homework. Rather than "give a proof..." say "how do I prove...." –  Qiaochu Yuan Apr 23 '11 at 22:37

1 Answer 1

Combinatorial proof:

Adapt the combinatorial proof for the Vandermonde identity by kissing one of the chosen men.

Proof by generating functions:

Adapt the proof for the Vandermonde identity by differentiating both sides.

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Kissing? Men? Not president of the committee? –  Mitch Apr 24 '11 at 1:25

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