Three straight lines l1, l2 and l3 are parallel and lie in the same plane. Five points are taken on each of l1, l2 and l3. The maximum number of triangles which can be obtained with vertices at these points, is

Question

Three straight lines l1, l2 and l3 are parallel and lie in the same plane. Five points are taken on each of l1, l2 and l3. The maximum number of triangles which can be obtained with vertices at these points, is (A) 425 (B) 405 (C) 415 (D) 505

Tardigrade Admin · Accepted Answer

The total number of points is 15 . From these 15 points we can obtain

^{15} C_{3}

triangles. However, if all the 3 points are chosen on the same straight line, we do not get a triangle. Therefore, the required number of triangles

=^{15} C_{3} - 3 (^{5} C_{3}) = 425.

Q. Three straight lines l1​,l2​ and l3​ are parallel and lie in the same plane. Five points are taken on each of l1​, l2​ and l3​. The maximum number of triangles which can be obtained with vertices at these points, is

425

405

415

505

The total number of points is 15 . From these 15 points we can obtain 15C3​ triangles. However, if all the 3 points are chosen on the same straight line, we do not get a triangle. Therefore, the required number of triangles =15C3​−3(5C3​)=425.

Q. Three straight lines $l_{1}, l_{2}$ and $l_{3}$ are parallel and lie in the same plane. Five points are taken on each of $l_{1}$ , $l_{2}$ and $l_{3}$ . The maximum number of triangles which can be obtained with vertices at these points, is

The total number of points is 15 . From these 15 points we can obtain $^{15} C_{3}$ triangles. However, if all the 3 points are chosen on the same straight line, we do not get a triangle. Therefore, the required number of triangles
$=^{15} C_{3} - 3 (^{5} C_{3}) = 425.$