First slide

Combinations

Question

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

Moderate

A

425
B

405
C

415
D

505

Solution

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

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