TITLE: Proving a certain binomial identity with three parameters
QUESTION [13 upvotes]: I would like to prove the following identity:
$$\sum_{m\geq 0} (-1)^{i-m}{m+k \choose m} {i-1 \choose m-1}{m+k+1 \choose j}  = \sum_{m\geq 0} {m+k \choose k}{k+1 \choose i-m}{k+1 \choose j-m}$$
for fixed $i,j \geq 1$, $k\geq 0$.  If it helps, I have a combinatorial interpretation of the RHS:  it counts the number of arrangements of $i$ $a$'s, $j$ $b$'s, and $k$ $c$'s so that the substrings "aa", "bb" cannot occur (there is no restriction on the c's.)  This I can show, although I have not found a direct combinatorial proof.  If it's helpful I can give my proof for the right-hand side; I derived it from the so-called Carlitz-Scoville-Vaughan theorem which I found out about recently on Math Overflow.
It is related to this question; I have a proof for it, contingent on proving this binomial identity.  I "stumbled" across this through some very optimistic guessing, but I'm not sure how to prove it.  I thought of using WZ theory, but as far as I know that only applies to identities with one parameter, not more general ones - but I'd be very happy if an algorithmic verification was possible.
Other possibilities:  Interpret the left-hand side as an inclusion-exclusion argument, or a determinantal formula.  Or, find a recurrence that both sides satisfy.

REPLY [7 votes]: Multiply both sides by $x^iy^jz^k$ and sum over all indices.
First, let's look at the left side:
$$
\begin{align}
&\sum_{m,i,j,k}(-1)^{i-m}\binom{m+k}{m}\binom{i-1}{m-1}\binom{m+k+1}{j}x^iy^jz^k\\
&=\sum_{m,i,j,k}\binom{m+k}{k}\binom{-m}{i-m}\binom{m+k+1}{j}x^iy^jz^k\\
&=\sum_{m,j,k}\binom{m+k}{k}\binom{m+k+1}{j}\left(\frac{x}{1+x}\right)^my^jz^k\\
&=\sum_{m,k}\binom{m+k}{k}\left(\frac{x}{1+x}\right)^m(1+y)^{m+k+1}z^k\\
&=\sum_{m,k}(-1)^k\binom{-m-1}{k}\left(\frac{x}{1+x}\right)^m(1+y)^{m+k+1}z^k\\
&=\sum_{m}\left(\frac{x}{1+x}\right)^m(1+y)^{m+1}(1-(1+y)z)^{-m-1}\\
&=\frac{1+y}{1-(1+y)z}\frac{1}{1-\frac{x}{1+x}\frac{1+y}{1-(1+y)z}}\\
&=\frac{(1+x)(1+y)}{(1+x)+(1+y)-(1+x)(1+y)(1+z)}\tag{1}
\end{align}
$$
Next, let's look at the right side:
$$
\begin{align}
&\sum_{m,i,j,k}\binom{m+k}{k}\binom{k+1}{i-m}\binom{k+1}{j-m}x^iy^jz^k\\
&=\sum_{m,j,k}\binom{m+k}{k}\binom{k+1}{j-m}x^m(1+x)^{k+1}y^jz^k\\
&=\sum_{m,k}\binom{m+k}{k}x^m(1+x)^{k+1}y^m(1+y)^{k+1}z^k\\
&=(1+x)(1+y)\sum_{m,k}(-1)^k\binom{-m-1}{k}(xy)^m((1+x)(1+y)z)^k\\
&=(1+x)(1+y)\sum_{m}\frac{(xy)^m}{(1-(1+x)(1+y)z)^{m+1}}\\
&=\frac{(1+x)(1+y)}{1-(1+x)(1+y)z}\frac{1}{1-\frac{xy}{1-(1+x)(1+y)z}}\\
&=\frac{(1+x)(1+y)}{(1+x)+(1+y)-(1+x)(1+y)(1+z)}\tag{2}
\end{align}
$$
Comparing $(1)$ and $(2)$ shows that the left and right hand sides are equal.