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Question 63

There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side AB, excluding A and B, of a triangle ABC. Similarly there are 6 points $$P_6, P_7,\ldots, P_{11}$$ on the side BC and 7 points $$P_{12}, P_{13},\ldots, P_{18}$$ on the side CA of the triangle. The number of triangles, that can be formed using the points $$P_1, P_2,\ldots, P_{18}$$ as vertices, is:

Total points: 5 on AB, 6 on BC, 7 on CA = 18 points total.

Total triangles from 18 points = C(18,3) = 816

Subtract collinear cases (3 points on same side don't form a triangle):

C(5,3) + C(6,3) + C(7,3) = 10 + 20 + 35 = 65

Number of valid triangles = 816 - 65 = 751

The correct answer is Option 3: 751.

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