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Bunuel - 3C8 shouldn't it be 8C3 ?
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Bunuel - 3C8 shouldn't it be 8C3 ?

It's just a different way of writing the same: choosing 3 out of 8 (it can not possibly be choosing 8 out of 3, right?).
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So if a polygon has n sides, the number of triangles to be formed will be nC3? Thanks!
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Bunuel - yes...my bad
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how many it can be in pentagon? 5c3?
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how many it can be in pentagon? 5c3?

True, because there are no set of 3 collinear points in a pentagon.

If there are n non-collinear points, where n>=3, we can make \(C^n_3\) triangles out of it.

If there is one set of x collinear points out of n points, where x>=3 and n>=x, then we can make

\(C^n_3-C^x_3\) triangles.



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