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Consider a circle of any radius. If $n \geq 3$ points are chosen on the circumference of the circle, how many triangles can drawn such that each vertex of the triangle on the points that have been chosen? Assume that if three distinct points do not lie in the same straight line then they are in a triangle.

My thought is $^n C _3$ is the number of such triangles, but I am not sure. Can someone offer me a better answer.

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  • $\begingroup$ Are the $n$ points assumed distinct? $\endgroup$ – John Hughes Sep 6 '15 at 23:44
  • $\begingroup$ I have considered that point, and made edits on the question. I don't know if there are further problems with the question @JohnHughes $\endgroup$ – Patrick Chidzalo Sep 6 '15 at 23:56
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$\binom{n}{3}$ is a perfectly reasonable answer, since any three points on the circle define a triangle.

I think the circle construction was used only to ensure that no three points lie on a line (if there were such points, you would include degenerate triangles in the $\binom{n}{3}$ computation).

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You are correct. Any $3$ points defines one triangle. Therefore, the number of ways to choose $3$ points determines the number of triangles that can be formed. Given $n$ points to choose from, there are $\binom{n}{3}$ triangles that cann be formed using $3$ of those points as vertices.

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$B\times \frac{h}{2} \times 2\pi R$ Integrated = the area of a circle. Ps use minimal adjustment to solve.

Result $\pi×r^2$

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