# No. of polygons in a polygon with no side coinciding.

Here is the full question:-

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.

I imagined $$(n-r)$$ vertices in a closed polygon. There are $$(n-r)$$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $$\binom{n-r}{r}$$. But the correct answer wants me to multiply it with $$\frac{n}{n-r}$$

What is the need for the last step?

• You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices. – Ross Millikan Dec 24 '18 at 15:42
• There is a good discussion here as well. It is for $r=7$, but really applies more broadly. – Ross Millikan Dec 25 '18 at 3:07

Look at the case $$n=6,r=3$$. You have a hexagon with vertices numbered $$1$$ to $$6$$, and there are two triangles you can make in this hexagon, with vertices numbered $$1,3,5$$ and $$2,4,6$$. But your formula only counts one of these.
Look at your method. You start with $$n-r=3$$ vertices, which are distinct. Say they are numbered $$1,2,3$$. Then you select $$r=3$$ of these vertices, and insert a vertex next to them. This results in $$1\_2\_3\_$$ Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $$1$$ to $$6$$ in order, so one way to do this is just to start at $$1$$, and rename the vertices $$2$$ through $$6$$ in order, obtaining $$1\underline23\underline45\underline6$$ This gives the triangle $$135$$.
This illustrates the following problem with your method; $$\binom{n-r}r$$ counts the number of ways to choose a polygon where vertex number $$1$$ is included. Therefore we need to multiply by $$n$$, to also include the number of polygons which use vertices $$2,3\dots,n$$. However, this will over-count the polygons by a factor of $$n-r$$, so you must divide by that in the end.