Question

how many line segments can be drawn through six points such that no three of them are collinear?
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Answer 1

Answer:

So, let's name the points A, B, C, D, E & F. i.e. 15 line segments. A, B, C, and D are four consecutive points such that no three points are collinear .

Answer 2

Answer:

The answer is 15

Given problem:  

How many line segments can be drawn through six points such that no three of them are collinear?  

Step-by-step explanation:

Given number of points = 6

Here no three points are collinear  

then the number of collinear points among 6 points will be 2

Note: Number of line segments can be formed with 'm' points in which 'n' points are collinear =  $\frac{m(m-1)}{2} - \frac{n(n-1)}{2} +1$  

Therefore, number of line segments can be formed with '6' points with '2' collinear points  =  $\frac{6(6-1)}{2} - \frac{2(2-1)}{2} +1$  

                           =    $\frac{6(5)}{2}-\frac{2(1)}{2} +1$    

                           =  3(5) - 1 + 1  = 15