To expand on Chetan4u's second method, each triangle can be formed with one point from each of the 3 lines AB (3 points), BC (4 points) and CA (5 points), or with two points from one line, and the third point from either of the other two lines.

The calculations look like this:

Scenario 1) One point from each line = \(3*4*5 = 60\)

Scenario 2) Two points from one line and one from the other two lines = \(3c2*9 + 4c2*8 + 5c2*7 = 27+48+70=145\)

(Quick explanation here in case I just lost some of you - Taking 2 points from the line AB, which has 3 points on it can be done in 3c2 ways - choosing two points from the 3 available on AB. Then we can choose the third point from any of the remaining points on the other two lines BC or CA, which have a total of 9 points among them. So the number of triangles that can be formed with two points on line AB is \(3c2*9 = 27\). Similarly for the other two lines.)

Add up the two scenarios: \(60+145 = 205\)

Answer: B

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Dave de Koos

GMAT aficionado