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I am working with a binomial sum that arises in some combinatorial arguments (and also appears in certain generating‐function manipulations). Specifically, I have this identity

$$ \sum_{j=0}^{n} \binom{k}{j}\,\binom{k}{n-j} \;=\; \sum_{j=0}^{n} \,(-1)^{\,n-j} \,\frac{k-n} {\displaystyle k - j}\displaystyle \binom{n}{j}\,\binom{k-1}{n}\,\binom{j+k}{n} \,\;=\; \binom{2k}{n}\ $$

I’m trying to understand why these two sums are equal from a combinatorial perspective, and also which binomial-coefficient identities or transformations are used to rewrite one side into the other.

Is there a known binomial identity that takes us directly from one to the other, or do I need a more intricate counting argument? Any insights or references to a standard identity would be really helpful.

If it's a new one please propose a name for it

Thank you everyone

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  • $\begingroup$ Why is it important to link the two sums rather than link the second sum directly to the binomial coefficient $\binom{2k}{n-1}$ ? $\endgroup$ Commented Mar 4 at 15:04
  • $\begingroup$ The goal here is to understand the combinatorial reasoning behind the equality of the two sums rather than just relying on an algebraic identity. If there is a direct counting argument that transforms one sum into the other, it would provide deeper insight into the underlying structure of the identity. Linking the second sum directly to $ \binom{2k}{n-1}\ $ might give a shortcut but wouldn’t necessarily reveal the combinatorial intuition behind it. $\endgroup$ Commented Mar 4 at 15:11
  • $\begingroup$ I want to know if the right side is a new identity, before starting to write a chapter about this in my article $\endgroup$ Commented Mar 4 at 15:19
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    $\begingroup$ Maybe more natural to rewrite slightly as $$\sum_{j=0}^{n} \binom{k}{j}\binom{k}{n-j} = \sum_{j=0}^{n} (-1)^{n-j} \frac{k-n}{k-j}\displaystyle \binom{n}{j} \binom{k}{n}\binom{j+k}{n}$$ $\endgroup$ Commented Mar 4 at 16:57
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    $\begingroup$ About your question whether the right identity is new. You can check this by writing it in standard hypergeometric notation. The sum is a multiple of $${}_3F_2\left(\begin{matrix}1-n,k+1,-k\\2+k-n,1-k\end{matrix};1\right).$$ This is a special case of the Pfaff-Saalschutz summation, see Eq. 16.4.3 on dlmf.nist.gov/16.4#ii. $\endgroup$ Commented Mar 4 at 18:12

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The identity is a special case of Saalschütz's theorem. The second sum may be written as $$(-1)^n \binom{k-1}{n}^2{}_3F_2\left({-n,-k,k+1\atop -k+1,k+1-n} \biggm|1\right)$$ and the $_3F_2$ can be evaluated by Saalschütz's theorem.

Added later: I missed Hjalmar Rosengren's comment, in which he said the same thing earlier.

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I want to know if the right side is a new identity, before starting to write a chapter about this in my article

This identity is known to Mathematica:

enter image description here

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  • $\begingroup$ Yes Mathematica can simplify it to their factorial form, so the identity is True Sum[Binomial[k, j]*Binomial[k, (n - 1) - j], {j, 0, n - 1}] == -Sum[(k*(-1)^(n - j)*Binomial[n - 1, j]* Binomial[k - 1, n - 1]*Binomial[j + k, n - 1])/(k - j), {j, 0, n - 1}] return true But it does not imply that the identity is new or not right ? Thank you very much for your comment $\endgroup$ Commented Mar 4 at 16:32
  • $\begingroup$ @Fibonacci : Right. However, I suspect that Mathematica derived this identity using some rather simple manipulations based on a vast collection of binomial identities that Mathematica has. $\endgroup$ Commented Mar 4 at 17:32
  • $\begingroup$ Yes totally agree with you, It can be nice if the method FullSimplify show the "stacktrace" $\endgroup$ Commented Mar 4 at 22:34
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    $\begingroup$ @IosifPinelis, unlikely. Searching for applicable identities is hard: Gosper's algorithm is easier and more general. $\endgroup$ Commented Mar 5 at 8:17
  • $\begingroup$ @PeterTaylor : Thank you for your comment. However, the accepted answer suggests that, at least in this case, the identity was derived simply based on a known family of identities. $\endgroup$ Commented Mar 5 at 15:52
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I arrived late to this game. Let's prove the identity with the help of the so-called Wilf-Zeilberger method of automatic proof.

To this end, divide through by the factor $\binom{2k}n$ so that we show the two identities \begin{align*} \sum_{j=0}^{n} \frac{\binom{k}{j}\binom{k}{n-j}}{\binom{2k}n}=1 \qquad \text{and} \qquad \sum_{j=0}^{n} (-1)^{n-j} \frac{(k-n)\binom{n}{j}\binom{k}{n}\binom{j+k}{n}}{(k-j)\binom{2k}n}=1. \end{align*} Define the functions $F_1(n,j):=\frac{\binom{k}{j}\binom{k}{n-j}}{\binom{2k}n}$ and $G_1(n,j):= - \frac12 \frac{\binom{k-1}{j-1}\binom{k}{n+1-j}}{\binom{2k-1}n}$. One can routinely check that $F_1(n+1,j)-F_1(n,j)=G_1(n,j+1)-G_1(n,j)$. Then, sum both sides over all integers $j$ (actually these are finite sums). The immediate impact is: the right-hand side vanishes. So, we obtain that $f_1(n+1)=f_1(n)$ where $f_1(n):=\sum_{j=0}^n F_1(n,j)$. Compute the value at, say $n=0$ (and get $f_1(0)=1$) to confirm the first promised identity.

I'll leave the 2nd identity because the procedure is very similar.

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