Question

Two straight line intersect at a point $O.$ Points $A_1,A_2,...A_n$ are taken on a line and points $B_1,B_2,...B_n$ are taken on the other line. If the point $O$ is not to be used, then number of triangles that can be drawn using these points as vertices, is

• A. $n(n-1)$
• B. $n(n-1{)}^2$
• C. $n^2(n-1)$
• D. $n^2(n-1{)}^2$

Answer 1

The correct option is C $n^2(n-1)$

$\because n$ points are taken at one line and another $n$ points are taken at other lines.
$\therefore$ Total $2n$ points are there in the plane in which $n$ points are in a line and another $n$ point are also in a line.

Hence, number of triangles $={}^{2n}{C}_3-{}^n{C}_3-{}^n{C}_3$
$=\frac{2n(2n-1)(2n-2)}{6}-\frac{2n(n-1)(n-2)}{6}$
$=\frac{1}{3}n(n-1)\left(3n\right)=n^2(n-1)$