How should I solve combination addition like this?

Count 6-subsets of $\{1,\dots,13\}$ by conditioning on the third smallest element $k$, which must be at least 3 and at most 10. For example, if $k=5$, then there are $\binom{4}{2}$ ways to choose two smaller elements from $\{1,\dots,4\}$ and $\binom{8}{3}$ ways to choose three larger elements from $\{6,\dots,13\}$. This combinatorial proof shows that $$\sum_{k=3}^{10} \binom{k-1}{2}\binom{13-k}{3}=\binom{13}{6}.$$

More generally, Identity 137 in Proofs That Really Count is: $$\sum_{j=r}^{n+r-k} \binom{j-1}{r-1}\binom{n-j}{k-r}=\binom{n}{k},$$ and the same combinatorial proof is given.