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?
Best Answer
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.