Number of triangles sharing all vertices but no sides with a given octagon 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?
If two of the vertices are $A$ and $C$, what are the possible third vertex?  Look at the whole list $A,...,H$
See bro, what I did, from 8 vertices, three will make a triangle... So 8C3 = 56 but sides are not to be included and one side will be forming a side of triangle so deducting 8 from 56 we will get 48 which is the answer

I hope this concept it true... Thanks