Number of ways to form hexagons by joining vertices of a convex 20-gon, such that no side of a hexagon is common with a side of the polygon? 
If hexagons are formed by joining the vertices of a 20 sided convex polygon such that no side of  hexagon is common with side of the polygon, then how many possible ways are there?

I have no idea where to start. I actually have difficulties when I deal with geometric applications of permutations and combinations. 
My approach: Choose 1 point (20 ways), then leave two adjacent points and choose 1 out of the remaining 17 (17 ways) and so on. But I think I can't use the multiplication principle without double counting and even if I double count, I can't find the method to remove those cases.
Would someone please help me to solve this question?

Imagine each of the points glued to a spacer that ensures that the space clockwise from the point is free. Each such point/spacer pair takes up $2$ slots, leaving $20-6\cdot2=8$ other slots. Thus we have two types of objects, $6$ of one and $8$ of the other, which we can linearly arrange in $\binom{14}6$ ways. Number the vertices starting from an arbitrary vertex, and apply these linear arrangements according to the numbers. But then we're not counting the arrangements where a point/spacer pair straddles the initial and final vertex. That's a fraction $\frac6{20}=\frac3{10}$ of the arrangements, so in total we have
$$
\frac1{1-\frac3{10}}\cdot\binom{14}6=\frac{10}7\cdot\binom{14}6=4290
$$
different hexagons.
