It is well known that
$$ \sum\_{k=1}^{n} k = \frac{n(n+1)}{2} .$$
As the story goes, Gauss notices that there are $n/2$ pairs of numbers that add up to $n+1$, hence the formula above.

But obviously the right hand side is ${n+1 \choose 2}$, namely
$$ \sum\_{k=1}^{n} k = {n+1 \choose 2} .$$
Is there a "combinatorial proof" of this second equation? I am trying to see the connection between the sum (and Gauss' method) to the problem of choosing $2$ objects from $n+1$ objects.