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Prove the identity $\binom{2n}{2}$ = $\binom{n}{2}+\binom{n}{n-2}+n^2$, where $n\geq2$, using a combinatorial proof.

I've tried to think of it in terms of a counting problem. I think that for the left hand side that given a group of people of size 2n you'll be choosing 2 people out of that. I think that the right hand side may be breaking down the left hand side into cases, but I'm not sure.

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  • $\begingroup$ I've tried to think of it in terms of a counting problem. I think that for the left hand side that given a group of people of size $2n$ you'll be choosing 2 people out of that. I think that the right hand side may be breaking down the left hand side into cases, but I'm not sure. $\endgroup$ – ivyleaf57 Feb 14 at 8:35
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    $\begingroup$ Please edit the question to include this information, and briefly mention in what context you are trying to solve this problem. If you do, I'll upvote your question. $\endgroup$ – jvdhooft Feb 14 at 8:39
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Hint: Split the $2n$ elements into two halves of size $n$. What are the possible ways $2$ chosen elements can be distributed between the halves? How many ways are there to choose them consistently with each distribution?

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  • $\begingroup$ I think I understand where the $\binom{n}{2}$ and the $n^2$ come from, but I'm not sure about the $\binom{n}{n-2}$. $\endgroup$ – ivyleaf57 Feb 14 at 11:53
  • $\begingroup$ @ivyleaf57 Since choosing $k$ items out of $n$ is the same as (not) choosing $n-k$ out of $n$, we have ${n \choose 2} = {n \choose n-2}$. Either choose two items left, choose two items right or choose one from both sides. $\endgroup$ – jvdhooft Feb 14 at 12:42

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