How many different possible ways are there to divide 6 ppl into two groups of 3?

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  • 3 years ago
    Favorite Answer

    From the 6 people, choose 3 to be in the first group:

    C(6,3) = (6 x 5 x 4) / (3 x 2 x 1) = 20 ways

    Then the remaining 3 people will automatically go in the second group:

    C(3,3) = 1 ways

    So it would seem there are 20 ways to split them into 2 groups of 3. But the groups both have 3 members, so they aren't distinguishable. For example, if I put Anna, Bob and Carol in the first group and Dave, Edwina and Francisco in the second group, that would look identical to putting Dave, Edwina and Francisco in the first group and Anna, Bob and Carol in the second group, it's really the same grouping.

    So you need to divide by 2! (or just 2) to account for the ways to switch the two indistinguishable groups.

    Answer:

    10 ways

    Specifically:

    ABC and DEF

    ABD and CEF

    ABE and CDF

    ABF and CDE

    ACD and BEF

    ACE and BDF

    ACF and BDE

    ADE and BCF

    ADF and BCE

    AEF and BCD

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  • 3 years ago

    6 choose 3 = 6!/3!(6-3)! = (6x5x4)/(3x2x1) = 120/6 = 20. So there are 20 ways to choose one group of 3 from a total of 6 people, once we choose the first group of 3, the second group of 3 is given. So the final answer is also 20.

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  • cidyah
    Lv 7
    3 years ago

    6! / (3! 3! 2!)

    = 720 / 72

    = 10

    Here's another example o dividing 10 people into two groups of five each.

    https://gyazo.com/b002ddb39718f64fcf0fa35fa09ffe64

    https://math.stackexchange.com/questions/500544/10...

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