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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?

reyna
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Abhinay Pandey
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  • 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 03:07

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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.

Mike Earnest
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