Ways to create a quadrilateral by joining vertices of regular polygon with no common side to polygon 
How many ways are there to create a quadrilateral by joining vertices of a $n$- sided regular polygon with no common side to that polygon?

It's quite easy to solve for triangles for the same question, logic remains same, we need to choose $4$ vertices with none of them being consecutive, what I did is broke them into cases i.e.: {$n$ sided polygon}
$1)$ All $4$ consecutive - $n$ ways
$2)$ $3$ consecutive - $n (n-5)$ ways (draw a diagram and don't take the dots adjacent to chosen pair)
$3)\; 2+2$ consecutive - $n$ ways to choose the first pair, to choose the second, ways are $n-4-1=n-5$ so, total ways $\Rightarrow n(n-5)$
$4)\; 2$ consecutive strictly - $n$ ways, now to choose the other two, leaving the adjacent ones, we have $n-4$ left, so, $(n-6)(n-10)+2(n-8)$
According to me the answer is $\binom n4 - \{ n + n (n-5) + n(n-5) + ((n-6)(n-10)+(2)(n-8))n \}$ 
but for $n=20$, our teacher told us that answer is near to $200$ can somebody please confirm my or my teacher's answer?
 A: Looking clockwise, attach one "general" vertex to each of the $4$ "special" vertices that will form the quadrilateral, viz, $\fbox{SG}$
Now there are $4$ boxes + $(n-8) = (n-4)$ entities.
Place the boxes in $\binom{n-4}{4}$ ways,
but as you actually have $n$, not $(n-4)$ vertices, multiply by $\frac{n}{n-4}$ to get
formula $=\frac{n}{n-4}\times\binom{n-4}{4}$ 
PS
Using black bullets for unused vertices, and white ones for quadrilateral vertices, create $4$ boxes $\boxed{\bullet\circ}\;$, each box to be treated as one unit. 
With $n=20$, say, the $4$ boxes can be placed anywhere in a string of $16$ units.  A diagrammatic representation (in a straight line)is given below.
$\bullet\bullet\boxed{\bullet\circ}\bullet\bullet\boxed{\bullet\circ}\bullet\bullet\bullet\boxed{\bullet\circ}\bullet\boxed{\bullet\circ}\bullet\bullet\bullet\bullet$
The basic idea is that any such arrangement prevents adjacent vertices from being used.
A: $1)$ All $4$ consecutive $=n$ is correct
$2)$ $3$ consecutive $=n(n-5)$ is also correct
$3)$ $2\times2$ consecutive $=\frac{n(n-5)}{2}$ You were double-counting in this case $(a,b,d,e) = (d,e,a,b)$
$4)$ $2\times1\times1=\frac{n(n-6)(n-5)}{2}$ [edited - this expression is cleaner]
Collectively:
\begin{align}&=n+2n\frac{(n-5)}{2}+n\frac{(n-5)}{2}+n(n-6)\frac{(n-5)}{2}\\&=n\left(1+(2+1+(n-6))\frac{(n-5)}{2}\right)\\&=n\left(1+(n-3)\frac{(n-5)}{2}\right)\\&=n\left(\frac{2+(n-3)(n-5)}{2}\right)\\&=n\left(\frac{n^2-8n+17}{2}\right)\\&=\frac{n^3-8n^2+17n}{2}\end{align}
The final answer is:$${n\choose4}-\frac{n^3-8n^2+17n}{2}$$
Confirm: $n=8\implies2; n=20\implies2275$ (Your teacher said near $2000$, right?)
