# 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.

Number of r-polygons in an n-polygon with no side coinciding
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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 $\left(n-r\right)$ vertices in a closed polygon. There are $\left(n-r\right)$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $\left(\genfrac{}{}{0}{}{n-r}{r}\right)$. But the correct answer wants me to multiply it with $\frac{n}{n-r}$. What is the need for the last step?
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benyaep17
Step 1
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 $1\mathrm{_}2\mathrm{_}3\mathrm{_}$
Step 2
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 $1\underset{_}{2}3\underset{_}{4}5\underset{_}{6}$
This gives the triangle 135.
This illustrates the following problem with your method; $\left(\genfrac{}{}{0}{}{n-r}{r}\right)$ ounts 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.
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