# 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?

• true blue anil's answer is much better than mine, being generalized to $\frac{n}{n-k}\times{n-k\choose k}$ for counting the number of k-gons that can be created from the vertices of an n-gon such that the k-gon and n-gon have no common side. – Logophobic Jan 29 '16 at 15:04
• @Logophobic: Thank you for your gracious comments. – true blue anil Jan 29 '16 at 15:51

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.

• sorry to ask, but a simple diagram would clarify it tremendously – Mrigank Jan 29 '16 at 15:41
• I have added a diagram, but am incapable of depicting a convex polygon, so have shown it in a straight line. You can sketch it out properly. – true blue anil Jan 29 '16 at 17:49
• True ball anil Nice explanation, Would you explain me the line, We have $n$ not $(n-4)$ virtices, So multiply by $\displaystyle \frac{n}{n-4}$, Thanks – juantheron Feb 20 '16 at 8:11
• @juantheron: true blue anil, thank you. By creating boxes, we reduced the number of entities (= vertices in this problem) from $n$ to $n-4$. So each entity has only $(n-4)$ starting points, but it should have $n$ starting points, hence the multiplication factor. – true blue anil Feb 20 '16 at 10:43

$1)$ 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?)

• sorry, had sleep deprivation .. in 4th, I wanted to do something like firstly take the points at ends of valid space i.e. 2 are possible and each would have (n-6) choices and all others possiblities-> (n-6) would have n-something, cases overlapped – Mrigank Jan 28 '16 at 21:14
• @ELiT then you wanted $\frac{n(n-6)(n-5)}{2}$, which is equivalent to what I originally posted as $4)$ -- I have reworked and edited my answer. – Logophobic Jan 28 '16 at 23:33