<s>[INST] Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$. 
A000332 has the comment:

Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.

The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$): 
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
 [/INST] Here is a combinatorial proof, with generalization.  Let $m$, $n$, and $r$ be non-negative integers such that $m\leq n$.  Consider $(m+r)$-subsets of the set $S=\{1,2,\ldots,n,n+1,n+2,\ldots,n+r\}$ which contain all numbers $n+1$, $n+2$, $\ldots$, $n+r$.  Such a subset must be of the form $V\cup \{n+1,n+2,\ldots,n+r\}$ with $V\subseteq \{1,2,\ldots,n\}$ having $m$ elements.  So there are $\binom{n}{m}$ possible subsets of this form.  We now count the number of these subsets in a different way, using PIE.
Let $T_s$ denote the set of $s$-subsets of $T=\{n+1,n+2,\ldots,n+r\}$.  Let $E_A$ denote the set of $(m+r)$-subsets $U$ of $S$ such that $A$ does not intersect $U$.  Let $\mathcal{C}$ be the set of all $(m+r)$-subsets of $S$.  Observe that $\mathcal{C}\setminus \bigcup_{k=1}^r E_{\{n+k\}}$ is the set composed by all $(m+r)$-subsets of $S$ which contain $n+1$, $n+2$, $\ldots$, $n+r$.  By PIE,
$$\left|\mathcal{C}\setminus \bigcup_{k=1}^rE_{\{n+k\}}\right|=\sum_{s=0}^r(-1)^s\sum_{A\in T_s}\left|E_A\right|.$$
Since $|E_A|=\binom{n+r-s}{m+r}$ and $|T_s|=\binom{r}{s}$, we get
$$\binom{n}{m}=\left|\mathcal{C}\setminus \bigcup_{k=1}^rE_{\{n+k\}}\right|=\sum_{s=0}^r(-1)^s\binom{r}{s}\binom{n+r-s}{m+r}.$$
In particular, when $m=1$ and $r=3$, we have the identity
\begin{align}n&=\binom{3}{0}\binom{n+3}{4}-\binom{3}{1}\binom{n+2}{4}+\binom{3}{2}\binom{n+1}{4}-\binom{3}{3}\binom{n}{4}\\&=\binom{n+3}{4}-3\binom{n+2}{4}+3\binom{n+1}{4}-\binom{n}{4}.\end{align}
This is equivalent to
$$\binom{n+3}{4}=n+3\binom{n+2}{4}-3\binom{n+1}{4}+\binom{n}{4}.$$
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