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There a 11 points on a plane with 5 lying on one straight line and another 5 lying on a second straight line which is parallel to the first line. The remaining point is not collinear with any two of the previous 10 points. How many triangles can be formed with the vertices chosen from these 11 points?

a) 85
b) 105
c) 125
d) 145

I calculated the number of triangles as follows: Taken one point on the first line, it can form 5 triangles with the non-collinear point and the 5 points on its opposite parallel line. So, the 5 points on the first line can form 5 x 5 = 25 triangles. Similarly, the points on the second line can form 25 triangles. Therefore, the 10 points on both the lines can form 25 x 25 = 125 triangles.

But, the answer to the question is 145 triangles. Where did I go wrong in calculating the number?

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First consider the triangles with two vertices on the first line. There are $10$ possible pairs of points on this line, so that they can form $60$ triangles ($50$ with the third vertex on the second line and $10$ with the third vertex on the non-collinear point).

Now consider the triangles with two vertices on the second line. By the symmetry of the problem, these are $60$ as well.

Finally, consider the triangles that do not have two vertices on the same line. These triangles clearly have one vertex on the first line, one vertex on the second line, and the third one on the non-collinear point. Because we can chose any of the $5$ points on each line, the number of these triangles is $5^2=25$.

Thus, the total number is $60+60+25=145$.

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