# How many cases can draw diagonals?

Imagine a n_regular polygon that vertex is named by 1 to n. We know can draw (n)(n+3)/2 diagonals in n_regular polygon,Also know if we want to draw Maximum diagonals that not intersecting each other inside the n_regular polygon is n−3 diagonals. For example for Flat hexagon we can draw Maximum 3 diagonals that we can do it with 14 cases.(you can find it easy) Now imagine we want to draw diagonals that have 2 condition like this: 1- any vertex is not connected to 2 before and 2 after vertex. 2- any 3 vertex is not exist that intersecting two by two inside the n_regular polygon. It's easy to understand that the Maximum of diagonals can draw is 2(n−5). But the question is how many cases can draw diagonals that Applicable 2 above condition? For n=6 we can draw 3 cases that necessitate this 2 condition . If we labeled regular when n=7 we have 2(7−5)=4 diagonals and we can not use this vertexes {1,4},{2,5},{4,7},{3,6} because we ignore the second condition.

Suppose we have a regular $n$-gon with vertices labeled 1 to $n$. There are $n(n-1)/2$ possible diagonals. The maximum number of non-intersecting diagonals is $n-3$, and if we draw them we get a triangulation. The number of ways to triangulate an $n$-gon is the Catalan number $C_{n-2}$. For example, the number of ways to triangulate a hexagon is $C_4 = \tfrac{1}{5} {8 \choose 4} = 14$.