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Three noncollinear points determine a triangle. How many triangles can be formed with 8 points, no three of which are collinear? A. 56 B. 24 C. 336 D. 6720
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u have to form subsets of three points out of the 8 given so u have \[ \large \binom{8}{3} \] triangles
Yup. Exactly as ^ said. \[nCr = \frac{n!}{(n - r)!r!}\]\[8C3 = \frac{8!}{(8 - 3)!3!}\]Can you finish it up?
no
harder
Well, the following is as far as I can get you... \[8C3 = \frac{\cancel{8!}}{\cancel{5!}3!} \implies \frac{8 \times 7 \times 6 }{3 \times 2 } \implies?\]
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56
Yup :)
thanks
np :)
This was posted an hour ago but I'm just now receiving a message.
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lol
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