A diagonal for a polygon is defined as the line segment joining two non-adjacent points. Given an n-sided polygon, how many different diagonals can be drawn for this polygon?
I know that the number of diagonals is C(n,2). However, I don't know how to account for the fact that you can't draw a diagonal for two points that are adjacent to each other.
There are two easy ways to take that restriction into account. You can start with the number of pairs of vertices, $\binom{n}2$, and subtract the number of pairs of adjacent vertices, $n$. Or you can observe that at each vertex there are $n-3$ other vertices that are not adjacent. That means that each vertex is an end of $n-3$ diagonals, so there are $n(n-3)$ ends of diagonals and therefore $\frac{n(n-3)}2$ diagonals.
As a check,
$$\binom{n}2-n=\frac{n(n-1)}2-n=\frac{n(n-3)}2\;.$$