Combinatorial proof that two ways of counting are equal: $\sum_{r=2}^{n+1} \binom r2 = \binom{n+2}3$

Question

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.

The right hand side is the number of 3 element subsets of $\{1,2,\ldots,n+2\}$. Maybe there is some other way of counting these subsets... — kimchi lover, Aug 12 '17 at 00:47
Also relevant and more general: [Proof of the Hockey-Stick identity](https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1) noting that for $r=0$ and $r=1$ you would have $\binom{r}{2}=0$ so can be added into the sum without changing it. — JMoravitz, Aug 12 '17 at 00:53
Related posts: [Want help for prove formula by combinatoric argument $1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$](https://math.stackexchange.com/q/1916505) and [Simplify triangular sum of triangular numbers: $\sum_{i=1}^{n}(\frac12i(i+1))$](https://math.stackexchange.com/q/1642906). — Martin Sleziak, Aug 12 '17 at 04:40

score 6 · Accepted Answer · answered Aug 12 '17 at 00:52

6

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.

answered Aug 12 '17 at 00:52

JMoravitz

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Thanks for your help, I got it almost immediately after thinking about what you said :) – Reece Jocumsen Aug 12 '17 at 01:35

Combinatorial proof that two ways of counting are equal: $\sum_{r=2}^{n+1} \binom r2 = \binom{n+2}3$

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