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Three straight lines \mathrm{l_1, l_2} and \mathrm{l_3} are parallel and lie in the same plane. Five points are taken on each of \mathrm{l_1, l_2} and \mathrm{l_3}. The maximum number of triangles which can be obtained with vertices at these points, is

Option: 1

425


Option: 2

405


Option: 3

415


Option: 4

505


Answers (1)

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The total number of points is 15 . From these 15 points we can obtain \mathrm{{ }^{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

\mathrm{={ }^{15} C_{3}-3\left({ }^{5} C_{3}\right)=425 }.

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