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There are 10 points in a plane and 4 of them are collinear. Find the number of triangles formed by producing the lines resulting from joining the points infinitely in both directions (assuming no two lines are parallel).

I can see that there are 10C2-4C2=40 straight lines. If no two pairs of lines were concurrent we would have 40C3=9880 triangles. However, I do not know how to adjust for the concurrent lines. Any help/suggestions much appreciated.

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In the statement, "assuming no two lines are parallel" probably means that the $$\binom{10}{2}-\binom{4}{2}+1=40$$ lines are distinct and pairwise concurrent. Then the number of triangles is $\binom{40}{3}=9880$.

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Actually, I think I was on the right track. The number of triangles eliminated for each of the 4 collinear points is 7C3=35 and the number of triangles eliminated from the 6 non-collinear points is 9C3=84. Hence we eliminate 4*35+6*84=644 triangles, resulting in 9880-644=9236 triangles formed.

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