Is Pascal's Rule the only such identity? I am wondering if Pascal's Rule is the only such identity in which
$$\binom{a}{b}+\dbinom{c}{d} = \binom{n}{m}$$
Precision aside on the expression above, in general I am wondering if there exist non-trivial examples where two binomial coefficients add to become another binomial coefficient. Any example with $b=1,d=1$ or $m=1$ is trivial: $$\binom ab + \binom cd = \binom{\binom ab + \binom cd}1\\\binom{a}b+\binom{\binom nm-\binom ab}1=\binom nm$$
Also, we can require $2b\le a$, $2d\le c$, and $2m\le n$ without loss of generality.
If there do exist non-trivial examples, how could one generate them?
 A: A relation:
${n \choose 2} + {{n \choose 2} \choose 2} = {{n \choose 2}+1 \choose 2}$
Non trivial, but quite obvious, because ${{n \choose 2} + 1 \choose 2} - {{n \choose 2} \choose 2} = {{n \choose 2} \choose 1}$ by Pascal's relation, and ${{n \choose 2} \choose 1} = {n \choose 2}$.
This actually generalizes to:
${n \choose p} + {{n \choose p} \choose 2} = {{n \choose p}+1 \choose 2}$
because ${{n \choose p } \choose 1} = {n \choose p}$.
Note that:

*

*$\forall n, {a_1n+a_2 \choose p} + {b_1n+b_2 \choose q} = {c_1n+c_2 \choose r}$, to be possible when $n \to \infty$, requires $p=r$ and (wlog) $q \le p$.

*$\forall n, {a_1n+a_2 \choose p} + {b_1n+b_2 \choose p} = {c_1n+c_2  \choose p}$ is impossible as soon as $p \ge 3$, because of Fermat-Wiles on $n^p$ coefficient.

So the only left case for ${a_1n+a_2 \choose p} + {b_1n+b_2 \choose q} = {c_1n+c_2 \choose r}$ is $p=r, q<r$ (wlog, as $p$ and $q$ can be switched).
For $p=r=3, q=2$, the only solutions I find are trivial ones - notably Pascal's relation.
