Thursday, 3 August 2017

combinatorics - Multiple sum involving binomial factors

Let n and m be positive integers and let 0jnm1. Show that:
\begin{align} \sum\limits_{l=m}^{n-j-1} \binom{n-l-1}{j} \binom{l}{m} \binom{n+l}{j} &=\sum\limits_{p=0}^j \sum\limits_{p_1=0}^j \sum\limits_{p_2=0}^m \frac{(p+p_1+p_2)!}{p! p_1! p_2!} \binom{j}{p_1} \binom{2n-j}{j-p}\\\hspace{1cm} &\times\binom{n-j+p}{m-p_2} \binom{n-j-m+p+p_2}{1+p+p_1+p_2} (-1)^{p+p_2} \end{align}




I have derived it using methods from analysis and then I verified the result using Mathematica. This result is a generalisation of another result given in here A double sum with combinatorial factors .

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