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I am looking for a closed form for the sum below

$$ \sum_{k=0}^{r} \frac{1}{2-\delta_{k, r-k}} \binom{n}{k} \binom{n}{r-k} $$

where $\delta_{k, r-k}$ is the Kronecker delta. I'd appreciate any helps in finding a closed-form solution for this sum.

To my understanding, this is basically equivalent to finding a closed-form solution to the following problem:

Given a set $S$ with size $|S|=n$, what is the number of two-element subsets of $\mathcal{P}(S)$, $\{A, B\} \in \mathcal{P}(S)$ such that $|A|+|B| = r$ for a fixed $r$.

I tried to use Wolfram Alpha, but it is given me this for the case where we count ordered pairs $(A, B)$ instead of sets (i.e. the sum if we ignore $\frac{1}{2-\delta_{k, r-k}}$), which contains factorial of a negative integer(!!) (and I don't know how to specify in Wolfram that $n>r>0$ and both are integers):

$$ \frac{(-1)^r (-2n+r-1)!}{(-2n-1)!r!} $$

nara
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    Are you wanting to count the number of ordered pairs $(A,B)$ with the property that $|A|+|B|=r$? Or are you wanting to count the number of sets ${A,B}$ with the property that $|A|+|B|=r$. Your summation counts the ordered pairs, not the sets. If wanting to count the number of sets, then you have overcounted each situation where $A$ and $B$ are nonequal twice. Further, if wanting to count the number of sets, then wouldn't you consider ${A,B}$ to be a one element set in the case that $A=B$? – JMoravitz Aug 09 '19 at 13:15
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    Have a look at Vandermonde's identity. – drhab Aug 09 '19 at 13:16
  • @JMoravitz Thanks for pointing it out. I guess either of them would work for me, as I'd just divide the answer for ordered pairs by 2 and add half of the product for $k=r-k$ case to get the answer for sets, but I actually want the answer for sets. I corrected the post accordingly (please do point it out if I didn't!). – nara Aug 09 '19 at 13:34
  • @drhab I think that's the answer – nara Aug 09 '19 at 13:34
  • @ArnaudD. Since apparently what I was looking for was just the Vandermondt identity, I think it may be a duplicate. – nara Aug 09 '19 at 13:35

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