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Three parallel lines lie in the same plane. If each line contains n marked points, what is the greatest possible number of triangles that can be formed using these points as vertices?
- 3n²(n−1)+1
- 3n²(n−1)
- n²(4n−3)
- None of these
Correct answer: n²(4n−3)
Solution
Choose any 3 of the 3n points: C(3n,3). Subtract the collinear triples on each of the 3 lines: 3*C(n,3). This equals [27n^3-27n^2+6n - (3n^3-9n^2+6n)]/6 = (24n^3-18n^2)/6 = n^2(4n-3).
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