alternating sums of terms of the Vandermonde identity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:30:54Z http://mathoverflow.net/feeds/question/8426 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8426/alternating-sums-of-terms-of-the-vandermonde-identity alternating sums of terms of the Vandermonde identity jk7xl8 2009-12-10T06:04:42Z 2009-12-10T09:09:07Z <p>Using Vandermonde's identity we know:</p> <p>$\sum_{i=0}^k \binom{k}{i}\binom{n-k}{n/2-i} = \binom{n}{n/2}$.</p> <p>I'm interested in how close the alternating sum is to 0 when k &lt;&lt; n. I.e., $\sum_{i=0}^k (-1)^i\binom{k}{i}\binom{n-k}{n/2-i}$.</p> http://mathoverflow.net/questions/8426/alternating-sums-of-terms-of-the-vandermonde-identity/8431#8431 Answer by Gjergji Zaimi for alternating sums of terms of the Vandermonde identity Gjergji Zaimi 2009-12-10T07:35:14Z 2009-12-10T09:09:07Z <p>So, you are interested in $f(n,k)=\sum_{i=0}^k (-1)^i\binom{k}{i}\binom{2n-k}{n-i}$. Simple manipulations show $f(n,k)=\frac{k!(2n-k)!}{(n!)^2}\left[\sum_{i=0}^n (-1)^i \binom{n}{i}\binom{n}{k-i}\right]$ Now the second factor counts the coefficient of $x^k$ in $(1-x^2)^n$ and therefore if $k$ is odd $f=0$ otherwise $f=(-1)^{\frac{k}{2}}\frac{k!(2n-k)!}{n!(k/2)!(n-k/2)!}$ which is far from zero...</p> <p>EDIT: On a different note I see the result is a signed generalized Catalan number of degree 2 (I was not aware they satisfied such simple identities). Since usually providing combinatorial interpretations for generalized Catalan numbers is not easy, may I ask in what combinatorial context did you face the above calculation?</p>