Question

Four points lie in a plane so that no three of them lie on a line. If lines are drawn connecting all pairs of these points, how many such lines are there?

Answer 1

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 $r$ elements, taken at a time, among $n$ elements are

.

So, the required number of lines

$=\binom{4}{2}=\frac{4!}{2!\times(4-2)!}=\frac{4!}{2!\times2!}=\frac{4\times3\times2\times\1}{2\times1\times2\times1}=6$

All the 6 possible lines can be verified from the figure.