Question

There are 11 points in a plane. No three of these lie in the same straight line except 4 points which are collinear. Find the number of triangles that can be formed for which the points are their vertices?

Answer 1

Number of points in a plane = 11

No three of these points lie in the same straight line except 4 points.

The number of triangles that can be formed for which the points are their vertices.

To form a triangle we need 3 non-collinear points.

We have the following possibilities.

(a) If we take one point from 4 collinear points and 2 from the remaining 7 points and join them.

The number of ways of selecting one point from the 4 collinear points is = ⁴C₁ ways

The number of ways of selecting 2 points from the remaining 7 points is = ⁷C₂

The total number of triangles obtained in this case is = ⁴C₁ × ⁷C₂

∴ The total number of triangles obtained in this case is

= ⁴C₁ × ⁷C₂ 

= `4 xx (7)/(2!(7 - 2)!)`

= `4 xx (7!)/(2! xx 5!)`

= `4 xx  (7 xx 6 xx 5!)/(2 xx 1 xx 5!)`

= 4 × 7 × 3

= 84

(b) If we select two points from the 4 collinear points and one point from the remaining 7 points then the number of triangles formed is

= ⁴C₂ × ⁷C₁ 

= `(4!)/(2!(4 - 2)!) xx 7`

= `(4!)/(2! xx 2!) xx 7`

= `(4 xx 3 xx 2!)/(2! xx 2!) xx 7` 

= `(4 xx 3 xx 7)/(2 xx 1)`

= 2 × 3 × 7

= 42

(c) If we select all the three points from the 7 points then the number of triangles formed is
= ⁷C₃ 

= `(7!)/(3!(7 - 3)!)`

= `(7!)/(3! xx 4!)`

= `(7 xx 6 xx 5 xx 4!)/(3! xx 4!)`

= `(7 xx 6 xx 5)/(3 xx 2 xx 1)`

= 7 × 5

= 35

∴ The total number of triangles formed are

= 84 + 42 + 35

= 161