# Binomial identity in Riordan's Combinatorial Identities

I am reading Riordan's Combinatorial Identities. On page 8 there is a formula like screenshot. I have no problem to understand the first three lines, in which we use $$(-1)^m \cdot \binom{-n}{m} = \binom{n + m - 1 }{m}$$ twice and Vandemonde identity in between. I could not understand the last step.

Note: It is helpful to use indices with lower and upper limit to clarify aspects of this kind.

We obtain \begin{align*} \color{blue}{\binom{n-p}{m}}&=(-1)^m\binom{p-n+m-1}{m}\tag{1}\\ &=(-1)^m\sum_{k=0}^m\binom{p}{m-k}\binom{-n+m-1}{k}\tag{2}\\ &=\sum_{k=0}^m\binom{p}{m-k}\binom{n-m+k}{k}(-1)^{m+k}\tag{3}\\ &\color{blue}{=\sum_{k=0}^m\binom{p}{k}\binom{n-k}{m-k}(-1)^k}\tag{4} \end{align*} and the claim follows according to your analysis and as indicated in the comment below.

Comment:

• In (1) we use the binomial identity $$\binom{p}{q}=\binom{-p+q-1}{q}(-1)^q$$.

• In (2) we apply Vandermonde's Identity*.

• In (3) we use again the binomial identity from (1).

• In (4) we shift the index $$k\to m-k$$.

I was stuck for a while then I realize I can just replace $$m-k$$ with $$u$$ and last step comes naturally from the previous step:

$$\binom{p}{u} \cdot \binom{n-u}{m-u}$$

The power of $$-1$$ becomes $$(-1)^{u+2k} = (-1)^u$$. Eventually rename $$u$$ with dummy variable $$k$$.