How many line segments can be determined by three given non-collinear points?.

Hint: In this question use the fact that a line segment is formed by minimum two points, so use the concept of combinations to select any 2 points from given three non-collinear points.

Complete step-by-step answer:

Let us consider A, B and C are the three given non-collinear points.
Now as we know a line segment can be drawn by using any two points but here the order is not important as line AB is the same as line BA.
So here we have to calculate combinations not permutations.
And we all know that a line segment requires two minimum points.
Therefore the combinations of number of line segments is ${}^3{C_2}$.
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property to expand the above combination we have,
$ \Rightarrow {}^3{C_2} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} = \dfrac{{3!}}{{2!.1!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} = 3$
Therefore the number of line segments from three non-collinear points is 3.
So this is the required answer.

Note: The trick point of this question was not the use of permutations and the relative arrangements of points plays no role as a line AB can’t be distinguished from line BA as only the naming is different. Non-collinear points mean that the points are not lying on the same straight line. These points are basically scattered in the x,y,z plane.