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Math Help - Group Combinations Problem

  1. #1
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    Group Combinations Problem

    If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

    Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

    Any thoughts?
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  2. #2
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    Quote Originally Posted by mathceleb View Post
    If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

    Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

    Any thoughts?
    I haven't worked this out rigorously, but I think the probability is \displaystyle \frac{1}{\binom{149}{2}}, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are \displaystyle \binom{149}{2} ways to choose two people out of the remaining 149...
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  3. #3
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    I'm not sure. I found this thread from back in the day which is a similar question...

    http://www.mathhelpforum.com/math-he...mutations.html
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  4. #4
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    Hello, mathceleb!

    We have a group of 150 people, broken into groups of 3, and only 3 are blue.
    What is the probability that all 3 blue people are put in the same group?

    The 150 people are divided into 50 groups of 3 people each.

    There are: . \dfrac{150!}{(3!)^{50}} possible groupings.


    To have the 3 blue people in one group, there is: . {3\choose3} = 1 way.

    The other 147 people are divided into 49 groups of 3: . \dfrac{147!}{(3!)^{49}} ways.
    . . Hence, there are: . \dfrac{147!}{(3!)^{49}} ways to have the 3 blue people in one group.


    The probability that all 3 blue people are in one group is:

    . . \dfrac{\dfrac{147!}{(3!)^{49}}} {\dfrac{150!}{(3!)^{50}}} \;=\;\dfrac{147!}{(3!)^{49}}\cdot\dfrac{(3!)^{50}}  {150!} \;=\;\dfrac{6}{150\cdot149\cdot148} \;=\;\dfrac{1}{551,\!300}

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  5. #5
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    Quote Originally Posted by undefined View Post
    the probability is \displaystyle \frac{1}{\binom{149}{2}}, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are \displaystyle \binom{149}{2} ways to choose two people out of the remaining 149...
    This correct.

    Quote Originally Posted by Soroban View Post
    [size=3]
    The 150 people are divided into 50 groups of 3 people each.
    There are: . \dfrac{150!}{(3!)^{50}} possible groupings.
    Those are ordered partions.

    There are \dfrac{150!}{(3!)^{50}(50!)} unordered partitions.

    There are \dfrac{147!}{(3!)^{49}(49!)} of those where the blues are together.

    Note that \dfrac{\dfrac{147!}{(3!)^{49}(49!)}}{ \dfrac{150!}{(3!)^{50}(50!)}}=\dfrac{1}{\binom{149  }{2}}
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  6. #6
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    Hmm since Soroban's answer differs from mine I decided to do a little more research.

    Quote Originally Posted by Soroban View Post

    There are: . \dfrac{150!}{(3!)^{50}} possible groupings.
    I believe this should be \dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!} because the groups of 3 can be permuted.

    Quote Originally Posted by Soroban View Post

    ...Hence, there are: . \dfrac{147!}{(3!)^{49}} ways to have the 3 blue people in one group.
    Likewise I believe this should be \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}.

    Leading to:

    \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)\cdot50  !}{49!\cdot\left(\dfrac{150!}{(3!)^{50}}\right)} = 50\left(\dfrac{1}{551,\!300}\right) = \dfrac{1}{11026} = \displaystyle \frac{1}{\binom{149}{2}}

    See here (a PDF file) for some more info; under subheading "Partitioning".

    Edit: Plato answered a lot quicker than I did, didn't refresh the page.
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