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anonymous

  • one year ago

math

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  1. Oleg3321
    • one year ago
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    well isnt it three students just go into each of the three groups so then all the nine students are split up evenly?

  2. Oleg3321
    • one year ago
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    did that help?

  3. anonymous
    • one year ago
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    I do not understand.. Hmm is it 9c3?

  4. zzr0ck3r
    • one year ago
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    9 choose 3

  5. zzr0ck3r
    • one year ago
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    \(\frac{9!}{3!(9-3)!}\)

  6. zzr0ck3r
    • one year ago
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    \(\dfrac{n!}{k!(n-k)!}\) tells you how many ways you can choose k things from n things.

  7. kropot72
    • one year ago
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    The number of ways 9 students can be partitioned into three teams containing 3 students each is found as follows: There are 9 choices for the first student of team A, 8 choices for the second student of team A and 7 choices for the third student of team A. The order of the choices does not matter. Therefore the number of ways of choosing team A is given by \[\large \frac{9\times8\times7}{3!}\] Having chosen team A, the number of ways of choosing team B is given by \[\large \frac{6\times5\times4}{3!}\] and the number of ways of choosing team C is given by \[\large \frac{3\times2\times1}{3!}\] Finally the total number of ways is given by \[\large \frac{9!}{3!3!3!}\]

  8. anonymous
    • one year ago
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    Thank you so much guys.. =)

  9. kropot72
    • one year ago
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    You're welcome :)

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