A rather curious identity on sums over triple binomial terms While exploring the Baxter sequences from my earlier MO post, I obtained a rather curious identity (not listed on OEIS either). I usually try to employ the Wilf-Zeilberger (WZ) algorithm to justify such claims. To my surprise, WZ offers two different recurrences for each side of this identity.
So, I would like to ask:

QUESTION. Is there a conceptual or combinatorial reason for the below equality?
$$\frac1n\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
=\frac2{n+2}\sum_{k=0}^{n-1}\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}.$$

Remark 1. Of course, one gets an alternative formulation for the Baxer sequences themselves:
$$\sum_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}}{\binom{n+1}1\binom{n+1}2}
=2\sum_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}}{\binom{n+1}1\binom{n+2}2}.$$
Remark 2. Yet, here is a restatement to help with combinatorial argument:
$$\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
=2\sum_{k=0}^{n-1}\binom{n+1}k\binom{n}k\binom{n+1}{k+2}.$$
 A: Just playing around with it: The RHS multiplied by $n$ is the same as
$$2 \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n}{k} \binom{n+1}{k+2}.$$
Subtracting this from $n$ times the LHS gives
$$\sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n+1}{k+2} \left( \binom{n}{k+1} - \binom{n}{k} \right).$$
Now you check that replacing $k$ with $n-1-k$ changes the sign of the summand, so the sum is zero.
A: The combinatorial identity
$$
\sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n+1}{k+1} \binom{n+1}{k+2} =
2 \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n}{k} \binom{n+1}{k+2}
$$
has a simple combinatorial proof.
The left-hand side counts all triples $A,B,C \subseteq [n+1]$ whose sizes are $k,k+1,k+2$ for some $k$. These come in two types:

*

*$n+1 \in B$. The triple $A,B\setminus\{n+1\},C$ has sizes $k,k,k+2$, and the middle set is a subset of $[n]$.

*$n+1 \notin B$. The triple $\overline{C},\overline{B}\setminus\{n+1\},\overline{A}$ has sizes $\bar{k},\bar{k},\bar{k}+2$, where $\bar{k}=n-1-k$, and the middle set is a subset of $[n]$.

The triples of each type are counted by the sum on the right-hand side.
