4. Inside a triangle (not on the sides) there are 3 points. Together with the 3 vertices of the original triangle, there are 6 points in total. Using these six points as vertices to form all non-overlapping triangles, if no three points among these 6 points are collinear, the number of all triangles is $n_{0}$; if three points among these 6 points are collinear (but no four points are collinear), the number of triangles formed is $n_{1}$; if four points among these 6 points are collinear, the number of triangles formed is $n_{2}$, then
(A) $n_{0}=n_{1}=n_{2}$;
(B) $n_{0}>n_{1}>n_{2}$;
(C) $n_{0}>n_{1} \geqslant n_{2}$;
(D) $n_{0} \geqslant n_{1}>n_{2}$.
4. (A)

Solution In any of these three cases, the number of non-overlapping triangles that can be formed with these 6 points as vertices is 7, i.e., $n_{0}=n_{1}=n_{2}=7$.