Number of $r$ - polygons in an $n$ - 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?
 A: 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.
