In a regular 12-sided polygon, how many triangles can be formed using vertices of the polygon that do not share a side with the polygon? I think is is 12 choose 3 minus 12, because that is the number of triangles that share a side with the polygon. Is that correct?

Another method, choosing the points of the triangle one by one then dividing by $6$ because each triangle has been chosen $6$ times (by choosing the vertices in different orders):
Choose the first vertex. $12$ options.
Choose the second vertex.
 - case (a) a next-but-one vertex: $2$ options
 - case (b) any other vertex: $7$ options
Choose the last vertex.
 - in case (a): $7$ options
 - in case (b): $6$ options
This gives $12 \times (2\times 7 + 7\times 6) = 672$ options, then the number of triangles is $ 672/6 = \fbox{112}$. 
The reason case (a) gives more selection options for the third vertex is that the next-but-one point shares a neighbour with the first point.
