permutation n combination

Question

There are 10 points in a plane ,no three of which are in the same straight line ,excepting  4 points ,which are  collinear.

find the 1)number of straight lines obtained from pairs of these points ;

            2)number of triangles that can be formed with the vertices as these points

Answer 1

1 : Number of Straight lines possible by joining 10 points in the plane (When no 3 or more than 3 points are collinear), selecting 2 points at a time    = ¹⁰C2  = 45

Similarly, Number of Straight lines possible by joining 4 points in the plane (When no 3 or more than 3 points are collinear), selecting 2 points at a time   = ⁴C₂  = 6

In the question, given 4 points are collinear, and since all the collinear points when joined pair wise form only One straight line,

Thus, required number of straight lines : = ¹⁰C₂ - ⁴C₂ + 1 = 40.

2 : The number of triangles that can be formed by 10 points (when no 3 points are collinear) = ¹⁰C₃.
Similarly, the number of triangles that can be formed by 4 points (when no 3 points are collinear) = ⁴C₃. 

In the question, given 4 points are collinear, and 4 collinear points cannot form a Triangle when taken 3 at a time.  ( With 3 or more Collinear points we cannot form any Triangle by joining any 3 of them.)

Thus, required number of triangle can be formed : = ¹⁰C₃ - ⁴C₃ = 120 - 4 = 116.