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How can we calculate the number of ways in which a $p$-sided polygon can be made inside a $n$-sided polygon such that no side of the $p$-sided polygon is in common with the $n$-sided polygon?

Matti P.
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imposter
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  • I can't really see how you would distinguish between the different ways. Isn't building a polygon, a continuous process? I mean, you can always slightly move one vertex ... Or am I missing something here? – Matti P. Jan 13 '20 at 07:10
  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jan 13 '20 at 07:12
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    I assume the vertices of the $p$-gon have to be vertices of the $n$-gon. Is that correct? Do you require the $p$-gon to be convex? [If so you presumably require the $n$-gon to be convex]. – almagest Jan 13 '20 at 09:14
  • As @joriki points out the OP would benefit from looking at the previous question but this question should not be said to be a duplicate just because a particular case has been asked previously. –  Jan 13 '20 at 10:27
  • @S.Dolan: I think that depends on whether the answer to the other question is easily generalizable. In this case it is. Everyone is free to post the general solution as a comment under the answer or edit it into the answer if it's deemed helpful to state the generalization explicitly; we don't need a separate question for that that's identical except for specific numbers plugged in. – joriki Jan 13 '20 at 17:27
  • There is no sense in which this OPs question involves "specific numbers being plugged in". –  Jan 13 '20 at 17:38

1 Answers1

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Pair each one of the $p$ vertices with the adjacent vertex in a clockwise direction. Then all we have to do is count the number of ways of selecting $p$ pairs of adjacent points from the $n$ available.

Let $v$ be one of the $n$ vertices.

Case 1 When $v$ is not one of the $p$ vertices we have to choose $p$ pairs from a line of $n$ points. We can think of this as the number of arrangements of $p$ pairs and $n-2p$ single points in a line.

Case 2 When $v$ is one of the $p$ vertices we have to choose a further $p-1$ pairs from a line of $n-2$ points.

The total is $\begin{pmatrix}n-p-1\\p-1\\\end{pmatrix}+\begin{pmatrix}n-p\\p\\\end{pmatrix}$.