Review on my method for $Number$ $of$ $diagonals$ in a regular $n$-gon is $\frac12n(n-3)$ I have an assignment on permutations and combinations topics. In that there is a question-
The number of interior angles of a regular polygon is $150^\circ$ each. The number of diagonals of the polygon is _____.
Attempt
I don't know about any thing about the P&C method for diagonals. So I found the side as per the formula $$(n-2)(360^\circ)=(\theta)n$$ where $n$ is the number of side and $\theta$ as interior angle.
But the main problem came to be the diagonal part because I was not able to understand the logic given on platforms like MSE itself. (Maybe it is due to language problem because I am still learning English language.) 
But I tried to count the polygon diagonal one by one to generate some pattern which may be useful. So I observed this-
In quads, if we start making diagonals then for first it will be $1$, then  again $1$ and $0$ and $0$.
In pents, the similar process gave $2$ then again $2$ then $1$ and then $0$, and $0$. 
In hexes, $3$ then $3$ then $2$ then $1$ then $0$ and again $0$.
Also, I observed that for the last two vertices it came to be $0$ always.
So observing this I first did in same way for hepts and after doing this I checked the answers for above cases in that formula $$\frac{n(n-3)}{2}$$. (Though I didn't understand how it was formed. And to my surprise, it came to be same. I even checked for $12, 13, 50, 40 \ldots$ It all gave the same value as that formula. 

Now my doubt is What I observed is right but how? Also I wrote it in terms of $$2(n-3)+(n-4)+...+1$$ to give $$\frac{n(n-3)}{2}$$. 


You can also use the Handshake Lemma from graph theory.  Let $G(V,E)$ be a graph on $n$ vertices, where the vertices  form a regular $n$-gon and the edges are the diagonals of the $n$-gon.  Then, prove that each vertex of $G$ has degree $n-3$.  By the Handshake Lemma, $G$ has
$$|E|=\frac{1}{2}\,\sum_{v\in V}\,\deg(v)=\frac{1}{2}\,n\,(n-3)=\frac{n(n-3)}{2}$$ edges.
