Question

Show that the number of diagonals of a polygon of n sides is $\frac{n(n-3)}{2}$.

Answer 1

The number of lines each joining 2 out of the n, points is
${}^n{C}_2$ = $\frac{n!}{2!(n-2)}$ = $\frac{n(n-1)}{2}.$
But these will include the n sides of the polygon.
Hence the number of diagonals is
$\frac{n(n-1)}{2}$ - n = $\frac{n(n-3)}{2}.$