MathDB
Prove this combinatorial formula

Source: 2009 József Wildt International Mathematical Competition

April 17, 2020
combinatoricsalgebra

Problem Statement

If n,p,qN,p<qn,p,q \in \mathbb{N}, p<q then ((p+q)nn)k=0n(1)k(nk)((p+q1)npnk)=((p+q)npn)k=0[n2](1)k(pnk)((qp)nn2k){{(p+q)n}\choose{n}} \sum \limits_{k=0}^n (-1)^k {{n}\choose{k}} {{(p+q-1)n}\choose{pn-k}}= {{(p+q)n}\choose{pn}} \sum \limits_{k=0}^{\left [\frac{n}{2} \right ]} (-1)^k {{pn}\choose{k}} {{(q-p)n}\choose{n-2k}}