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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?
Solution
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
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