Let m points be A1,A2,A3,...Am
We consider four points A1,A2,A3,A4
If these points are joined in all possible cases we have 3 points of intersection H1,H2andH3
So the required points of intersection =3.mC4
=3.m!4!(m−4)!=m!8(m−4)!
Hence, option 'A' is correct.