Talk:Vandermonde's identity

Chu-Vandermonde identity sum bound

For the Chu-Vandermonde identity the sum should be from k=0 to n not infinity right? --Ray andrew 02:23, 27 September 2007 (UTC)

The sum is finite only if n is an integer. If n is a general complex value, then the sum runs to infinity. There's no difference ... when n is an integer, all of the higher order terms are zero anyway. linas (talk) 15:03, 25 March 2008 (UTC)

Errr ... I believe that the identity should hold for general complex n, which would follow from Gausses hypergeometric function identity, although I have not tried to check that for any forehead-slapping gotcha's (e.g. some wacky violation of Carlson's theorem.) linas (talk) 15:10, 25 March 2008 (UTC)

Merge from Vandemonde's formula

I suppose that Vandemonde is a misspelling for Vandermonde. Based on http://mathworld.wolfram.com/VandermondesFormula.html it seems that it relates to the Chu-Vandermonde identity. If appropriate, please merge content. —Leo Laursen ( T ¦ C ) 17:27, 30 January 2008 (UTC)

There was nothing to merge, it was a trite restatement of the content of this page. I just turned the later into a redirect to this article. linas (talk) 15:01, 25 March 2008 (UTC)

Clarification required for Algebraic proof

I am a math BA student yet I could not understand how the following identity:

${n \choose k}={n! \over k!(n-k)!},$

Proves the Vandermonde's identity. however in the french page there is a much clearer proof using equality between polynomials. I suggested that unless there is a one-liner explanation, we will provide a more detailed proof here (again, the french page provides one which I could figure without the need to understand the language) Weiss gal wiki (talk) 14:35, 16 August 2008 (UTC)

Include a Sketch for the Geometric Proof?

The geometric proof had me cross-eyed until I sketched out a picture. Couldn't a diagram be included in this section? That way, it's easy to work out the length of all the segment sequences: as an example, the final, vertical sequence of segments for the first (and easiest to diagram) combination would be of length n - r + k, for any k. I think just showing the first (and simplest) combinations for the m part and the n part--where all the earliest segments are included (and therefore horizontal) and the latter segments are excluded (and therefore vertical)--would be enough for most readers to get an idea how the proof works. (To avoid confusion: the orientations in this description are particular to the sketch I made. "Horizontal" and "vertical", "included" and "excluded" could each be switched in another diagram and that one would still be applicable to the proof.) 115.87.197.31 (talk) 07:04, 11 July 2014 (UTC)Nick Palevsky

graphic as example

https://i.ibb.co/Rvzr7wT/gamble3copy-min.png --Backinstadiums (talk) 17:47, 15 November 2019 (UTC)