When I was reading an article by CHUN-GANG JI (A SIMPLE PROOF OF A CURIOUS CONGRUENCE BY ZHAO), he mentioned in the acknowledgement
the following identity


$$\sum\_{i+j+k=p,\text{ } i,j,k\gt 0}{p\choose i}{p\choose j}{p\choose k}={3p\choose p}-3{2p\choose p}+3$$


Can anybody give me a lead to a combinatorial or algebraic proof of this identity ?



A combinatorial identity involving binomial coefficients
Without the inequality on the $i, j, k$, the sum on the left would be $\binom{3p}{p}$. This is an immediate consequence of the Chu-Vandermonde convolution identity


$$\sum\_{i+j = k} \binom{x}{i}\binom{y}{j} = \binom{x+y}{k}$$


which is treated in many places, for example in Concrete Mathematics by Graham, Knuth, and Patashnik.


By an inclusion-exclusion argument, the adjustment one must make to account for the inequality in the sum is to subtract the sum where at least one of the $i, j, k$ is $0$, then add back in the sum where at least two of them are $0$. For example, the sum over all cases where $k = 0$ is


$$\sum\_{i+j=p} \binom{p}{i}\binom{p}{j} = \binom{2p}{p}$$


where again we use Chu-Vandermonde. This leads to the answer you want.


