Is the similarity between (sum of the first $n$ numbers) = $\frac{n(n+1)}{2}$ and $\binom{n}{2} = \frac{n(n-1)}{2}$ a coincidence? These formulae are equal to one another, except for an off-by-one shift. Is this a coincidence? The derivation of
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
has to do with pairing off elements. So are they related in any meaningful way?
 A: No, this is not a coincidence in the sense that there is a proof which explains why it should be true. This is a standard method known as "double counting".
Consider having $n$ people at a party, who have never met. We want to count the number of handshakes. Each pair of people only shake hands once.
One way to do this is to consider each individual person. The first person makes $n-1$ handshakes, then the second person makes $n-2$ handshakes, etc., down to the second to last person making $1$ handshake.
Another way is to do it combinatorically. We are looking for pairs of people without regard for order, that is, $\binom{n}{2}$. Hence:
$$ \binom{n}{2} = 1 + 2 + \cdots + (n-1) $$
A: I don’t think the “pairing off” of numbers in the usual proof of the sum has any connection to the two-element subsets counted by $n\choose2$, but here’s a way to see a connection between two-element subsets of a set of $n$ elements and the sum of the numbers $1$ (or zero) through $n-1$. The idea is similar to what George described in his answer.
Consider the set $S=\{0,1,\dots,n-1\}$. You want to know if it’s a coincidence that the sum of the numbers in $S$ equals the number of two-element subsets of $S$.
Each two-element subset of $S$ has a unique largest element (which is an element of $S$). There are no two-element subsets in which $0$ is the largest element. There is one two-element subset in which $1$ is the largest element, and so on. For each $k\in S$, there are $k$ two-element subsets in which $k$ is the largest element, and so on.
Stated another way, each number $k\in S$ appears as the largest value in exactly $k$ of the two-element subsets of $S$. 
Therefore, to count the number of two-element subsets of $S$, we can just add up the numbers in $S$, the result being $0+1+\cdots+(n-1)$.
