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catgirl17

  • 2 years ago

a four person committee is chosen at random from a group of 15 people. how many different committees are possible?

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  1. amistre64
    • 2 years ago
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    15 choose 4

  2. amistre64
    • 2 years ago
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    15!/(4! 11!)

  3. amistre64
    • 2 years ago
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    15.14.13.12.11! --------------- 4.3.2.11! 15.14.13.12 --------------- 4.3.2 15.7.13

  4. lgbasallote
    • 2 years ago
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    is this permutation? or combination @amistre64 ?

  5. amistre64
    • 2 years ago
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    combo since it doesnt suggest positions for the 4

  6. amistre64
    • 2 years ago
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    its just a group of 4 people with no specific need for position abc = bca in this case

  7. lgbasallote
    • 2 years ago
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    i see thanks

  8. catgirl17
    • 2 years ago
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    where did you get the 15x7x13 ?

  9. lgbasallote
    • 2 years ago
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    15.14.13.12 ---------- 4.3.2.1 15.14.13.6 ---------- 4.3. 15.14.13.2 --------- 4 15.14.13 -------- 2 15.7.13

  10. lgbasallote
    • 2 years ago
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    do you get it @catgirl17 ?

  11. catgirl17
    • 2 years ago
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    where did the 6 come from in step 2?

  12. ParthKohli
    • 2 years ago
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    \(\Large \color{MidnightBlue}{\Rightarrow nCr = {n! \over (n - r)!*r! } }\) Got it?

  13. ParthKohli
    • 2 years ago
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    nCr can be represented as: \(\Large \color{MidnightBlue}{ (\matrix{n \\r}) }\)

  14. catgirl17
    • 2 years ago
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    no i don't get it @lgbasallote

  15. lgbasallote
    • 2 years ago
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    hmm but you get 15.14.13.12 ----------- 4.3.2.1 right?

  16. catgirl17
    • 2 years ago
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    yes

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