# r-sided polygons are formed by joining the vertices of an n-sided polygon.

Find the number of polygons that can be formed, none of whose sides coincide with those of the n-sided polygon.

Since we have to form an r sided polygon, it is obvious that we would have to select r vertices from the n vertices of the n-sided polygon the restriction being, that no consecutive vertices should be chosen. So to ensure that these vertices are separated we can select $$n-r \choose r$$ vertices that are between the selected vertices as separators.
This is where I got stuck. I tried searching the web but couldn't find any satisfactory explanations. Could someone please help me out with the logical approach to solve this?

• Essentially a duplicate of math.stackexchange.com/q/1663877. They just do $r=7$, but the same logic applies to any $r$. Feb 22 at 15:16

Assuming you start with a convex polygon and you want to select convex polygons, you have to select $$r$$ vertices from $$n$$ where $$r \ge 3$$ and you do not select adjacent vertices.
Suppose one of the vertices is $$A$$. Then you want $$r-1$$ non-adjacent vertices from the $$n-3$$ vertices which are not $$A$$ or its neighbours.
You could use induction or a form of stars and bars to show this is $${n - r-1 \choose r-1}$$
But you need to multiply this by $$n$$ as the number of choices for $$A$$, and divide by $$r$$ as each polygon is counted multiple times, suggesting the answer may be $$\frac nr{n - r-1 \choose r-1}$$ at least when $$r \ge 3$$. As a check, this give the obvious answer of $$2$$ when $$n=2r$$.
• @PrajwalTiwari You have to select from $n-3$ vertices subject to the constraint that the $r-1$ selections are non-adjacent vertices: that reduces the number of possibilities. If you had been selecting $k$ non-adjacent vertices from $m$ there would have been ${m-k+1 \choose k}$ possibilities Feb 22 at 11:46