Question

There are three coplanar lines, if any m points taken on each of the lines, the maximum number of triangle with vertices at these points is

• A. $m^2(4m-3)$
• B. $3m^2(m-3)+1$
• C. $3m^2(m-3)$
• D. $m(4m-3)$

Answer 1

The correct option is A $m^2(4m-3)$

There are three case

C1: When all the points are on the same line then number o triangles are zero

C2: when $2$ points are on one line and third point is on the other lines

C3: When all the three points are on three different lines.

C1: $0$

C2: $3\times$${mC}_2\times$${2mC}_1$

C3: Choose one from all and multiply, it give us $m^3$ triangles.

Total number of triangles can be found by adding all the three cases

$m^2(4m-3)$