Relation between Combinatorics and Sum of First $(n-1)$ Natural Numbers

Question

Is it mere coincidence that ${n \choose 2}$ is equal to the sum of the first $(n-1)$ natural numbers both of which happen to be $\frac{n(n-1)}{2}$?

A Proof of the relation is appreciated. Thanks

Answer 1

It is not.

The number of two-element subsets of the set $S = \{1, 2, 3, \ldots, n\}$ is $$\binom{n}{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n(n - 1)}{2}$$

We can also count the set by considering two-element subsets with largest element $i$. The number of such subsets is $i - 1$ since there are $i - 1$ ways to choose the smaller element. Since $i$ can vary from $1$ to $n$, the number of two-element subsets is $$\sum_{i = 1}^{n} (i - 1) = \sum_{k = 0}^{n - 1} k = \frac{n(n - 1)}{2}$$ where we have made the substitution $k = i - 1$.

Answer 2

Ok so the best way to understand this is by visualizing a deck of $n$ cards, now, if you pick out the first card then you have $n-1$ ways to pair it with another card you pick out, and then if you pick out another then you have $n-2$ ways, and so on. You will get the series, $$n-1 + n-2 + n-3\ldots + 3+ 2+1.$$ Which is, $$\sum _{i=1}^{i=n-1} i$$ Also, try and think about what you are doing here, you are essentially finding the number of ways to pick $2$ cards out of a deck of $n$ cards. Which can be written as $$\binom{n}{2}$$ So you get, $$\binom{n}{2} = \sum _{i=1}^{i=n-1} i $$
Another way to think of it is the number of ways you can make subsets of size $2$ from an $n$ sized set, which is given as $\binom n 2$ namely $$A = \{x_1,x_2 + \ldots + x_{n-1} + x_n + \} \subseteq \mathbb{N}$$ where $$x_i \lt x_{i+1}$$ and a subset $$Q = \{a,b\}\subseteq \mathbb{N}$$ where $a<b$. Now, for $$b = x_n$$ we have $n-1$ options for $a$ and for $$b=x_{n-1}$$ there are $n-2$ options and so on until for $$b=x_1$$ there are $0$ options for $a$ so we get $$\sum _{i=1}^{i=n-1} i = \binom n 2$$