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On a regular hendecagon (2 dimensions), how many non-symmetric ways can you draw a maximal set of non-intersecting diagonals?

This puzzle has been driving me crazy! I've been doing it by hand.

-The maximal set is defined as the largest number of diagonals that do not intersect, except at the endpoints

-Please also note reflects and rotations are not valid. (i.e. only counts as one)

Any ideas of how to go about problem solving would be helpful!

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    I think the Catalan numbers are what you want except that rotations and reflections are considered different. I found it in http://oeis.org/A033282, where the right side of the triangle is the Catalan numbers.2012-08-20

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I'm not entirely sure about this, but I think you can count them like this:

For $8$ diagonals to fit in, there must be one that connects vertices that are only $2$ apart. Let's number the vertices consecutively such that two such vertices are numbered $2$ and $4$. Then either $2$ must be connected to $5$, or $4$ to $1$. Without loss of generality, assume that $4$ is connected to $1$. From now on, we always have a binary choice of drawing a line from either vertex of the line just drawn to the nearest neighbour of the other vertex. These choices can be associated with the $6$ central diagonals, which are either dead ends or connect two parts of the set. These six choices would yield $2^6$ sets, but the ones without mirror symmetry are counted twice. There are $2^3$ with mirror symmetry, so the total is $(2^6-2^3)/2+2^3=(64-8)/2+8=36$.

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    If this method of counting is correct, then [OEIS A005418](http://oeis.org/A005418) is the associated sequence, with $n=1$ corresponding to a pentagon.2012-08-20