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Sagot :
Solution:
Given that;
If you have 1 point spread out randomly, so that no 3 points are colinear
Colinear points are points that lines on the same line.
Drawing points randomly from the given 3 points
Applying the combination formula,
[tex]\begin{gathered} nCr=\frac{n!}{\left(n-r\right)!r!} \\ Where \\ n=3 \\ r=2 \\ 3C2=\frac{3!}{2!(3-2)!}=\frac{3!}{2!1!}=\frac{3\times2\times1}{2\times1\times1}=3 \end{gathered}[/tex]Hence, the number of lines you can draw is 3
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