Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.

Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.

[tex]\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.[/tex]

So far I agree with yma's answer. Here are the planes.

ABC

ABD

ABE

ACD

ACE

ADE

BCD

BCE

BDE

CDE

Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is

parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.

A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.

[tex]\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.[/tex]

What are those lines and in what planes are they found?

AB: ABC ABD ABE

AC: ABC ACD ACE

AD: ABD, ACD, ADE

AE: ABE, ACE, ADE

BC: ABC BCD BCE

BD: ABD BCD BDE

BE: ABE BCE BDE

CD: ACD BCD CDE

CE: ACE BCE DCE

DE: ADE BDE CDE

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