In how many ways can we select three vertices from a regular polygon having $2n+1$ sides ($n>0$) such that the resulting triangle contains the centre of the polygon?
HINT: When we fix a vertex of the $(2n+1)$-gon, we can assume that there are $n$ vertices on left of the fixed vertex and $n$ vertices on the right. In order for the resulting triangle to contain the center of the polygon, we need to choose three vertices such that when we fix every vertex one by one, we should have:
$1$ vertex on the left of fixed vertex, $1$ vertex on the right of fixed vertex (and in this case, the last vertex will be the fixed vertex) OR
$2$ vertices on the left, $1$ vertex on the right OR
$2$ vertices on the right, $1$ vertex on the left.
It might not be so clear with words so I will give an example from enneagon ($9$-gon with $n = 4$):
Suppose we choose the vertices $A$, $D$ and $E$. Then when we fix $C$, $A$ is on the right of $C$; $D$ and $E$ is on the left of $C$ so it may seem like argument holds (by second case). However, I said "fix every vertex one by one" so for example when we fix $A$, both $D$ and $E$ is on the left of $A$ therefore $\Delta ADE$ doesn't include the center. But, if we choose $A$, $F$ and $E$, whichever vertex we fix, we will see that one of the three cases above will hold. So $\Delta AFE$ includes the center.
Hint: Distribute the remaining $2n-2$ vertices to the $3$ gaps between the chosen vertices. Find the number of ways to distribute them such that each gap contains no more than $n-1$ vertices. (because if a gap contains $n$ or more vertices, the centre of the polygon would be outside the triangle.)