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.
How many pairs can we obtain with $n+1$ objects?
$1+2+3+4+\cdots+n\\ =\binom{1}{1}+\binom{2}{1}+\binom{3}1+\binom{4}1+\cdots+\binom{n}1\\ =\binom{2}{2}+\binom{2}1+\binom{3}1+\binom{4}1+\cdots+\binom{n}1$
and use $\binom{a}{b+1}+\binom{a}{b}=\binom{a+1}{b+1}$.
This may not be what you are looking for, but I think it's a little cool.
This can be viewed as an instance of Vandermonde's Identity
$$
\begin{align}
\sum_{k=1}^nk
&=\sum_{k\in\mathbb{Z}}\binom{k}{k-1}\binom{n-k}{n-k}\tag1\\
&=(-1)^{n-1}\sum_{k\in\mathbb{Z}}\binom{-2}{k-1}\binom{-1}{n-k}\tag2\\
&=(-1)^{n-1}\binom{-3}{n-1}\tag3\\[3pt]
&=\binom{n+1}{n-1}\tag4\\[3pt]
&=\binom{n+1}{2}\tag5
\end{align}
$$
Explanation:
$(1)$: $\binom{k}{k-1}=k\,[k\ge1]$ and $\binom{n-k}{n-k}=[k\le n]$
$(2)$: negative binomial coefficients
$(3)$: Vandermonde's Identity
$(4)$: negative binomial coefficients
$(5)$: symmetry of Pascal's Triangle
