Two straight lines intersects at a point O. Points A1, A2, ......... An are taken on one line and B1, B2, ........ Bn on the other. If the point O is not to be used, the number of triangles that can be drawn using these points as vertices, is :

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Two straight lines intersects at a point O. Points A₁, A₂, ......... A_n are taken on one line and B₁, B₂, ........ B_n on the other. If the point O is not to be used, the number of triangles that can be drawn using these points as vertices, is :

n²(n – 1)²

n(n – 1)

n²(n – 1)

n(n – 1)²

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Let AC and BD are two straight lines intersect at 0. Points A₁, A₂, …, A_n are taken on AC and B₁, B₂,…B_n on BD as shown:

So, there are n points
on AC as well as BD.
To draw a triangle without using 0, two points to be taken from one of the line as vertices and one point from the other line.

Total number of triangles = ⁿC₂ ⋅ ⁿC₁ ⋅ 2 (since any line can be chosen for two points and other for one)

$= \frac{n!}{2! \cdot (n - 2)!} \times \frac{n!}{(n - 1)!} \times 2$

= n(n – 1) . n

= n²(n – 1)