Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many (i) straight lines (ii) triangles can be formed by joining them?


Given that we need to find the no. of different straight lines that can be drawn from the 18 points in which 5 are collinear.


We know that 2 points are required to draw a line and the collinear points will lie on the same line, and only one line can be drawn by joining any two points of these collinear points.


Let us assume the no. of lines formed be N,


N = (total no. of lines formed by all 18 points) – (no. of lines formed by collinear points) + 1


Here 1 is added because only 1 line can be formed by the four collinear points.


N = 18C25C2 + 1


We know that ,


And also n! = (n)(n – 1)......2.1





N = 153 – 10 + 1


N = 144


The total no. of ways of different lines formed are 144.


Given that we need to find the no. of triangles that can be drawn from the 18 points in which 5 are collinear.


We know that 3 points are required to draw a triangle and the collinear points will lie on the same line, and no triangle can be drawn by joining any three points of these collinear points.


Let us assume the no. of triangles formed be N,


N = (total no. of triangles formed by all 18 points) – (no. of triangles formed by collinear points)


N = 18C35C3


We know that ,


And also n! = (n)(n – 1)......2.1





N = 816 – 10


N = 806


The total no. of triangles formed are 806.


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