Question

There are three coplanar parallel lines. If any n points are taken on each of the lines, the maximum number of triangles with vertices at these points will be

• A. 3n² (n –1)
• B. 3n² (n –1) + 1
• C. n² (4n –3)
• D. None of these

Answer 1

Answer: (C)

n² (4n –3)

Answer 2

The maximum no. of triangles = No. of triangles with vertices on different lines + No. of triangles with 2 vertices on one line and third vertex on any one of the remaining two lines

= ⁿC₁. ⁿC₁ .ⁿC₁ + ³C₁ . ⁿC₂. ²ⁿC₁

= n³ + 3 $\frac{n(n - 1)}{2}2n$= 4n³ – 3n²

= n² (4n –3)