Question

There are 11 points in a plane. No three of these lies in the same straight line except 4 points, which are collinear. Find,

(i) The number of straight lines that can be obtained from the pairs of these points?

(ii) The number of triangles that can be formed for which the points as their vertices?

Answer 1

(i) 4 points are collinear

Total number of points 11.

To get a line we need 2 points

∴ Number of lines = ¹¹C₂ = (11 x 10)/(2 x 1) = 55

But in that 4 points are collinear

∴ We have to subtract ⁴C₂ = (4 x 3)/(2 x 1) = 6 ... (2) 

From (1) Joining the 4 points we get 1 line 

∴ Number of lines = ¹¹C₂ – ⁴C₂ + 1 = 55 – 6 + 1 = 50

(ii) A triangle is obtained by joining 3 points. 

So selecting 3 from 11 points can be done in ¹¹C₃ = (11 x 10 x 9)/(3 x 2 x 1) = 165 

But of the 11 points 4 points are collinear. So we have to subtract ⁴C₃ = ⁴C₁ = 4 

∴ Number of triangles = 165 – 4 = 161