How many lines are determined by 8 points, no more than 3 of which are collinear? How many triangles are determined by the same points if no more than 4 are coplanar?

Collinear points

We have 8 points (and just to have them named, we'll have #A,B,C,D,E,F,G,H#)

A line is defined by 2 points, and so we can have lines #AB, AC, AD,...,GH#. But we are being limited by being told that no more than 3 points are on the same line.

Let's consider the case where we have #A,B,C# on the same line - that means that #AB, AC, BC# are all the same line. No other point can be on this line and since 2 points are needed to define a line, any other point (say for instance, #D#) can't be on a line with any two of #(A,B,C)#.

Let's go one further. Let's take line #DE#. It can run through #A#, or #B#, or #C#, but not any more than one of them (otherwise it'll be collinear with all three points, #A, B, C#). So let's assume that we have line #ADE#. We know points #B, C# are not collinear with this and just like before, any of the remaining points (#F, G, H#) can't be collinear with any two points #A, D, E#.

So what lines have we made so far that have 3 points collinear?

#ABC, ADE#

Following this pattern, what other lines with 3 collinear points can we make? As I'm working through this, I'm keeping in mind that the maximum number of times I'll be able to list each point is three:

#AFG, BDF, CDG, BEG, HCE#

I don't think I can make any more lines like this, so we have 7 with three collinear points.

And now let's make the remaining lines. These will be two point lines that haven't been connected by our three point lines:

#AH, BH, CF, DH, EF, FH, GH#

And so I count 14 lines in total.