5 points in 3D space: how many planes and planes intersections?

Question

Given 5 points in space such that no three of them are colinear and no four of them are coplanar. If we consider all the planes containing any 3 of these 5 points, and the intersections of all these planes taken two by two, how many lines will we obtain, at most?

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At first, my idea was just to say that there are $C(3,5) = 10$ (binomial coefficient) different planes. And then any combination of 2 planes must give a different intersection line: therefore $C(2,10)=45$ should be the answer?

However, I sense that this is naive and overly simple. Could I be missing something?

Answer 1

You are right that there are $\binom{5}{3}= \color{red}{10}$ planes BUT ...

The $3$ planes $(1,2,3),(1,2,4),(1,2,5)$ will intersect in the line $(1,2)$.

Can you fix things from here ?

Answer 2

OK, based on the past comments, here is how I see it:

There are 10 pairs of points, i.e. through each pair we can fit 3 planes (the planes going through that pair as well as each of the other 3 points separately). Indeed, then there is only 1 common intersection to these 3 planes. Therefore, the planes formed this way would have just 10 intersections that coincide with an existing pair of points (out of the theoretical 30 'intersections' we could form with the naive approach I first used).

There remains another 15 planes intersections which do not coincide with any existing pairs of points (these are the intersections of 123-145, 123-245, 123-345, 124-135, 124-235, 124-345, 125-134, 125-234, 125-345, 134-235, 134-245, 135-234, 135-245, 145-234, 145-235). Therefore, IF (but are we sure of this?) none of these intersection are also the same, we would have 10+15=25 total intersection for these 10 planes? Is that correct?

This is intriguing: is this a famous result which generalizes to N points?