# Why is the number of diagonals in an $n$-sided polygon $\frac{n\cdot(n-3)}{2}$?

A couple days into my math lessons I learned that the formula for finding the number of diagonals in polygon is

$N_d=\frac{n\cdot(n-3)}{2},$

where $N_d$ is the number of the diagonals and $n$ is the number of sides.

I think it is because in the polygon we can put diagonal line from each point to all other points, but not for the ones that are connected with the side of polygon to us. That's why the it is $n\cdot(n-2)$ and we are dividing by the because we are counting each diagonal 2 times.

Is my thinking correct?

• Yes, you have the right idea Mar 7, 2017 at 16:22
• That's pretty much a perfect reasoning. Mar 7, 2017 at 16:22
• I got it wrong, because there isn't diagonal from vertex A to vertex A. BTW: fixed. Mar 7, 2017 at 16:33

## 2 Answers

Diagonals join non-adjacent vertices. We have $n$ choices for the first vertex, but then we have $n-3$ choices for the second, as we cannot use adjacent vertices, nor can we choose the same vertex.. We then divide by two to account for direction - the diagonal joining A to B is the same as that joining B to A. Thus,

$$N_d = \frac{n(n-3)}{2}$$

with 2 points we have $1$ line ,with 3 poins we have $3$ lines ,with 4 points we have $6$ lines ...

with $n$ points we have $\binom n2$ lines

so

the number of diagonals in an nn-sided polygon $=$

(all of the lines) - $(n \space sides)=$ $$\binom n2-n=\\ \dfrac{n!}{2!(n-2)!}-n=\\ \dfrac{n(n-1)}{2}-n=\\ \dfrac{n(n-1)}{2}-\dfrac{2n}{2}=\\\dfrac{n(n-3)}{2}$$