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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{How many triangles can be formed from 6 given points on a circle} \hspace{.33em}\\~\\ & a.)\ 6! \hspace{.33em}\\~\\ & b.)\ 3! \hspace{.33em}\\~\\ & c.)\ \dfrac{6!}{3!} \hspace{.33em}\\~\\ & d.)\ \dfrac{5\times 6\times 4}{6} \hspace{.33em}\\~\\ \end{align}}\)

  2. Michele_Laino
    • one year ago
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    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

  3. welshfella
    • one year ago
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    so d?

  4. Michele_Laino
    • one year ago
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    yes!

  5. mathmath333
    • one year ago
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    answer given is d.)

  6. mathmath333
    • one year ago
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    thank vermuch

  7. Michele_Laino
    • one year ago
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    :)

  8. anonymous
    • one year ago
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    dern just missed it didn't I?

  9. dan815
    • one year ago
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    any 3 chosen points will make a triangle

  10. dan815
    • one year ago
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    you need 3 different points

  11. dan815
    • one year ago
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    so 6 choose 3

  12. mathmath333
    • one year ago
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    what if it asked quadrilaterals 6C4 ?

  13. dan815
    • one year ago
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    yep

  14. dan815
    • one year ago
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    just gotta be careful though

  15. dan815
    • one year ago
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    that the points arent colinear

  16. dan815
    • one year ago
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    it should say in this question too that no 3 points are collinear

  17. dan815
    • one year ago
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    oh wait it says on a circle xD

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