Finding the diagonals of an n-sided polygon

Question

I know this is elementary but bear with me.

Is there a way to calculate the number of diagonals an n-sided polygon has without using combinatorics???

I noticed that for every vertex there is an $\geq1$ other vertices to which it connects. For instance: In triangles it doesn't exist I guess.

In a quadrilateral $n=4$ a vertex connects to one other vertex.

A pentagon $n=5$. A vertex connects to $2$ other ones

Hexagon $n=6$: A vertex to $3$ other..

Can anything be made out of this, say a $1550$ sided polygon? Also what happens as $n\rightarrow\infty$

Answer 1

As you have noted, in an $n$-gon, each vertex shares a diagonal with $n-3$ other vertices (every vertex except for itself and its two neighbors). There are a total of $n$ vertices, so $n(n-3)$ diagonals.

Not quite!! This way we have double counted all of the diagonals. So the actual number is $\frac{n(n-3)}{2}$.

Answer 2

Each of the $n\choose 2$ pairs of vertices gives rise to a diagonal-or-edge. Subtract the $n$ edges to arrive at ${n\choose 2}-n=\frac{n(n-1)}2-n=\frac{n(n-3)}2$ diagonals.