Counting question

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align} & \normalsize \text{How many triangles can be formed from 6 given points on a circle} \hspace{.33em}\\~\\ & a.)\ 6! \hspace{.33em}\\~\\ & b.)\ 3! \hspace{.33em}\\~\\ & c.)\ \dfrac{6!}{3!} \hspace{.33em}\\~\\ & d.)\ \dfrac{5\times 6\times 4}{6} \hspace{.33em}\\~\\ \end{align}}\)
by induction, on the number of points, I think that we can get: \[\Large \left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right)\] triangles
so d?

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Other answers:

yes!
answer given is d.)
thank vermuch
:)
dern just missed it didn't I?
any 3 chosen points will make a triangle
you need 3 different points
so 6 choose 3
what if it asked quadrilaterals 6C4 ?
yep
just gotta be careful though
that the points arent colinear
it should say in this question too that no 3 points are collinear
oh wait it says on a circle xD

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