$${2\choose 2} {10\choose 3}+{3 \choose 2}{9 \choose 3}+{4 \choose 2}{8 \choose 3}+{5 \choose 2}{7 \choose 3}+{6 \choose 2}{6 \choose 3}+{7 \choose 2}{5 \choose 3}+{8 \choose 2}{4 \choose 3}+{9 \choose 2}{3 \choose 3}={13 \choose 6}$$
I am supposed to get the answer ${13\choose 6}$. I wonder if there is any formula for combination addition like this or are there some tricks to do 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.