[Question:]
There are three coplanar parallel lines. If any $p$ points are taken on each of the lines, the maximum number of triangles with vertices on these points is
[The solution given is:]
Select any three points from total $3p$ points, which can be done $^{3p}C_3$ ways.
But this also includes selection of three collinear points.
Now three collinear points from each straight line can be selected in $^pC_3$ ways.
Then the number of triangles is $^{3p}C_3−3(^pC_3)=p^2(4p−3)$.
The part I don't understand is would it not be possible for $3$ points selected form $3$ different lines to be collinear?
We have accounted for collinear points present on the same line through $3(^pC_3)$ but not the case where they are present on different lines.