There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points isa)3p2(p-1)+1b)3p2(p-1)c)p2(4p-3)d)none of theseCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

The number of triangles with vertices on one line and the third vertex on any one of the other two lines.

∴ the required number of triangles = p³ +3p²(p-1)
Note: The word “maximum” ensures that no selection of points from each of the three lines are collinear.

Answer

Explanation:

To find the maximum number of triangles with vertices at the given points, we need to consider the different combinations of points that can form triangles.

Let's consider the three coplanar parallel lines as line 1, line 2, and line 3.

Number of points on each line:
Let's assume there are p points on each line.

Choosing points for the first side of the triangle:
To form a triangle, we need to choose 3 points. Let's start by choosing points on line 1 for the first side of the triangle. We can choose any 3 points from the p points on line 1. The number of ways to choose 3 points from p points is given by the combination formula: C(p, 3) = p! / (3!(p-3)!).

Choosing points for the second side of the triangle:
For the second side of the triangle, we need to choose 2 points from the remaining p points on line 2. The number of ways to choose 2 points from p points is given by the combination formula: C(p, 2) = p! / (2!(p-2)!).

Choosing points for the third side of the triangle:
For the third side of the triangle, we need to choose 1 point from the remaining p points on line 3. The number of ways to choose 1 point from p points is given by the combination formula: C(p, 1) = p.

Total number of triangles:
To find the maximum number of triangles, we need to multiply the number of ways to choose points for each side of the triangle.

Number of triangles = C(p, 3) * C(p, 2) * C(p, 1)
= (p! / (3!(p-3)!) * (p! / (2!(p-2)!) * p
= p(p-1)(p-2)/3! * p(p-1)/2! * p
= p^3(p-1)^2 / 3! * 2! * 1
= p^3(p-1)^2 / 6
= (p^3(p-1)^2) / 6
= p^2(4p-3)

Hence, the maximum number of triangles with vertices at these points is p^2(4p-3).

Therefore, the correct answer is option 'C': p^2(4p-3).

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