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$A, B, C, D$ and $E$ are coplanar points and three of them lie in a straight line. What is the maximum number of triangles that can be drawn with these points as their vertices?
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The correct answer is:
9
Number of triangles using 5 points out of three are on a straight line $={ }^{5} C_{3}-{ }^{3} C_{3}=\frac{5 !}{3 ! 2 !}-1=\frac{5 \times 4}{2}-1$
$=10-1=9$
$=10-1=9$
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