Answer:
6
Step-by-step explanation:
Given that 4 points lie in a plane so that no 3 points of them lie on a line.
Let A, B, C, and D are four points as shown in the figure.
One line to be drawn, only 2 points are needed.
As no 3 points are collinear, so the number of combination of 2 points among the total 4 points gives the number of lines can be drawn.
As the total number of combinations of [tex]r[/tex] elements, taken at a time, among [tex]n[/tex] elements are
.
So, the required number of lines
[tex]=\binom{4}{2}=\frac{4!}{2!\times(4-2)!}=\frac{4!}{2!\times2!}=\frac{4\times3\times2\times\1}{2\times1\times2\times1}=6[/tex]
All the 6 possible lines can be verified from the figure.
