We know that by using any 3 non collinear points we can form a triangle, so number of triangles which can be formed by connecting \(n\) points such that no three points are collinear is given by \(^n{{\rm{C}}_3}\). 

Let us assume that out of \(n\) points, \(k\) are collinear then  these \(k\) points will not contribute any triangle. So the number of triangles will be \(^n{{\rm{C}}_3} - {\,^k}{{\rm{C}}_3}\)

In the same way we know that by joining any two points we get a straight line, so the number of straight line which can be formed joining \(n\) distinct points will be in \(^n{{\rm{C}}_2}\) provided that no three points are collinear. 

Let us assume that out of \(n\) points, \(k\) are collinear then these \(k\) points will contribute only one straight line, so the number of straight lines will be \(^n{{\rm{C}}_2} - {\,^k}{{\rm{C}}_2} + 1\)

Also every two lines intersect in one point, so number of intersection points which can be generated by \(n\) straight lines will be \(^n{{\rm{C}}_2}\) to provided that no three lines pass through the same intersection point.

Let us assume that out of \(n\) lines, \(k\) lines are passing through only one intersection point, then the number of intersection points will be \(^n{{\rm{C}}_2} - {\,^k}{{\rm{C}}_2} + 1\)

Example 01: There are 20 points in a plane out of which 10 are collinear, find the number of straight lines which can be formed by joining any 2 points.

Example 02: Find the total number of triangle in the given diagram.

In this triangle the points P, Q and R are not connected to all other points. The points A, B, C, D, E, F and G are connected to each other. Hence total number of triangles because of these 7 points = \({}^7{{\rm{C}}_3} - 6 \times \,{}^3{{\rm{C}}_3} = 35 - 6 = 29\). 

Now because of only point P there are 6 new triangles:

▲PCF, ▲PCE, ▲PFG, ▲PEG, ▲PFD and ▲PED.

Similarly the number of triangles because of Q and R is 12.

Number of triangle = 29 +18 = 47