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
Bookmarks