Prove:
$$\sum_{r=2}^{n+1} {r \choose 2} = {n+2 \choose 3}$$
I know that ${n+2 \choose 3}$ is ${n \choose 3}$ with repetition (${n + 3 - 1 \choose 3}$), but apart from that I don't even know what to do. Any help would be appreciated.
2026-03-29 12:32:05.1774787525
Combinatorial proof that two ways of counting are equal: $\sum_{r=2}^{n+1} \binom r2 = \binom{n+2}3$
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Hint: For each 3-element subset of $\{1,2,\dots,n+2\}$ there is a highest element and two more elements which are strictly smaller than it.