Two straight lines intersect at a point O. Points \[{A_1},{A_2}, \cdots ,{A_n}\] are taken on one line and points \[{B_1},{B_2}, \cdots ,{B_n}\] on the other. If the point O is not be used, then what is the number of triangles that can be drawn using these points as vertices?
A. \[n\left( {n - 1} \right)\]
B. \[n{\left( {n - 1} \right)^2}\]
C. \[{n^2}\left( {n - 1} \right)\]
D. \[{n^2}{\left( {n - 1} \right)^2}\]

Hint: There are two cases to solve the question. First case is we take one point from the first line and two points on the second line. Second case is we take two points from the first line and one point on the second line. Then solve both equation by using combination formula and add them.

Formula Used: Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step by step solution: To make a triangle we need three points. Here we will consider two cases.
First case: Taking two points from first line and one point from the second line to make a triangle.
Second case: Taking two points from first line and one point from the second line to make a triangle.
The number of points on first line is n. The number of points on second line is 2.
First case:
The number of ways of choosing two points from first line is \[{}^n{C_2}\].
The number of ways of choosing one point from second line is \[{}^n{C_1}\].
The number of triangles by choosing two points from first line and one point from second line is \[{}^n{C_2} \cdot {}^n{C_1}\].
\[ = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} \cdot \dfrac{{n!}}{{1!\left( {n - 1} \right)!}}\]
\[ = \dfrac{{n\left( {n - 1} \right) \cdot \left( {n - 2} \right)!}}{{2!\left( {n - 2} \right)!}} \cdot \dfrac{{n \cdot \left( {n - 1} \right)!}}{{1!\left( {n - 1} \right)!}}\]
\[ = \dfrac{{n\left( {n - 1} \right)}}{2} \cdot n\]
\[ = \dfrac{{{n^2}\left( {n - 1} \right)}}{2}\]
Second case
The number of ways of choosing one point from first line is \[{}^n{C_1}\].
The number of ways of choosing two points from second line is \[{}^n{C_2}\].
The number of triangles by choosing one point from first line and two points from second line is \[{}^n{C_1} \cdot {}^n{C_2}\].
\[ = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}} \cdot \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}\]
\[ = \dfrac{{n \cdot \left( {n - 1} \right)!}}{{1!\left( {n - 1} \right)!}} \cdot \dfrac{{n\left( {n - 1} \right) \cdot \left( {n - 2} \right)!}}{{2!\left( {n - 2} \right)!}}\]
\[ = n \cdot \dfrac{{n\left( {n - 1} \right)}}{2}\]
\[ = \dfrac{{{n^2}\left( {n - 1} \right)}}{2}\]
Thus total number of triangles that can be drawn using these points as vertices is \[ = \dfrac{{{n^2}\left( {n - 1} \right)}}{2} + \dfrac{{{n^2}\left( {n - 1} \right)}}{2}\]
\[ = {n^2}\left( {n - 1} \right)\]

Option ‘C’ is correct

Note: Student often do mistake to solve this question. They consider only first condition. They do not consider the second case. For this reason they are unable to reach the correct answer.