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$$ \sum_{k=1}^{2n} { 2n \choose k } { 2n \choose 2n-k } = \binom{4n}{2n} - 1$$

This equation have nice methods? Thank you everyone,

This problem is from this Proving $\binom{2n}{n}\ge\frac{2^{2n-1}}{\sqrt{n}}$

Community
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math110
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  • Kind of tough to do, since I think it is not true. Looks more like $\binom{4n}{2n}-1$. – André Nicolas Aug 12 '13 at 06:16
  • why not change the summation to start at $k=0$? – Harry Richman Oct 12 '17 at 20:24
  • actually, given the title of this question and the motivation I suspect that the original intent may have been to ask for a proof of the identity $\sum_k \binom{2k}{k} \binom{2(n-k)}{n-k} = 2^{2n}$ (which is also a duplicate https://math.stackexchange.com/questions/37971/identity-for-convolution-of-central-binomial-coefficients-sum-limits-k-0n) – Harry Richman Oct 12 '17 at 20:32

2 Answers2

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$$LHS+1=\text{coeff of $x^{2n}$ in} (1+x)^{2n}(1+x)^{2n}=\text{coeff of $x^{2n}$ in} (1+x)^{4n}={4n \choose 2n}\ne2^{4n}$$

Kunnysan
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Find instead the sum from $k=0$. There is a group of $2n$ men and $2n$ women. We want to choose $2n$ people. This can be done in $\binom{4n}{2n}$ ways.

Let us count the number of choices another way. We could choose $0$ men and $2n$ women, or $1$ man and $2n-1$ women, and so on. There are $\binom{2n}{k}\binom{2n}{2n-k}$ ways to choose $k$ men and $2n-k$ women. Add up.

For the version that starts from $k=1$, subtract the $1$ way to choose $0$ men and $2n$ women.

André Nicolas
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