The number of triangles whose vertices are the the vertices of the vertices of an octagon but none of whose sides happen to come from the sides of the octagon.
My Attempt: Let $\{A,B,C,D,E,F,G,H\}$ be the vertices of an octagon. It is given that none of the side of octagon is the side of the triangle, so we do not take consecutive points.
So we take either $\{A,C,E,G\}$ OR $\{B,D,F,H\}$ points out of which we will take only three points, because we have form a triangle.
So This can be done by $\binom{4}{3}+\binom{4}{3} = 8$
But the only options given are 24, 52, 48, and 16.
Where have I made a mistake?
Best Answer
If two of the vertices are $A$ and $C$, what are the possible third vertex? Look at the whole list $A,...,H$