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Math Help - Multinomial Coefficients: 2 Questions

  1. #1
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    Multinomial Coefficients: 2 Questions

    1. Twenty-five girls are bored.. How many ways may the girls form two teams of 5 to play basketball against each other?

    - This I used C(25,5) * C(20,5) and got 823,727,520. The answer in the back says it is 411,863,760. It might just be coincidence but the answer in the back is half the number I got.

    2. A fraternity consisting of 30 members wants to play 7 vs. 7 flag football. How many different match-ups are possible?

    -This I used C(30,7) * C(23,7) and got 4.99x10^11. The answer in the back says 2.495x10^11. Again, half my answer.
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  2. #2
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    Re: Multinomial Coefficients: 2 Questions

    Hello, kangta27!

    I hope I can explain this clearly.


    1. Twenty-five girls are bored. .How many ways may the girls
    form two teams of 5 to play basketball against each other?

    I used C(25,5)\cdot C(20,5) and got 823,\!727,\!520.
    The answer in the back says it is 411,863,760.
    It might just be coincidence but the answer in the back is half the number I got.

    Let's name the twenty-five girls: Ann, Beth, Carol, Doris, . . . Yvonne.
    Also, let's name the two teams: Reds and Blues.


    C(25,5) = 53,\!130 is the number of ways to select the Red Team.

    Then C(20,5) = 15,\!504 is the number of ways to select the Blue Team.

    The product (823,\!727,\!520) is the number of ways to form the two teams.


    Included in this long list are these two partitions:

    . . \begin{array}{cc}\text{Red} & \text{Blue} \\ ABCDE & FGHIJ \end{array} \qquad\begin{array}{cc} \text{Red} & \text{Blue} \\ FGHIJ & ABCDE \end{array}


    Since the two teams do not have names,
    . . these two cases represent the same division into two teams.

    That is, our list has twice as many partitions than actually exist.

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  3. #3
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    Re: Multinomial Coefficients: 2 Questions

    Thanks Soroban! So in these situations, does that mean I always divide the answer by 2?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Multinomial Coefficients: 2 Questions

    Suppose you have a collection of n objects and you wish to form 2 groups from the set each with m objects where 2m\le n.

    By the commutativity of multiplication, it doesn't matter if you choose m items to go into the first group, then m items to go into the second group, or if you alternate in any manner. By the fundamental counting principle, we find the number of permutations is:

    n(n-1)(n-2)\cdots(n-(2m-2))(n-(2m-1))

    Now we must account for the permutations within the two groups themselves:

    \frac{n(n-1)(n-2)\cdots(n-(2m-2))(n-(2m-1))}{m!m!}=

    \frac{n\cdots(n-(m-1))}{m!}\cdot\frac{(n-m)\cdots(n-(2m-1))}{m!}=

    \frac{n!}{(n-m)!m!}\cdot\frac{(n-m)!}{(n-2m)!m!}=

    {n \choose m}\cdot{n-m \choose m}

    Now we must account for the permutations of the number of groups:

    \frac{1}{2!}{n \choose m}\cdot{n-m \choose m}=

    \frac{1}{2}{n \choose m}\cdot{n-m \choose m}

    One could generalize to say that for p groups (where pm\le n) the number of ways N is:

    N=\frac{1}{p!}\prod_{k=0}^{p-1}{n-km \choose m}
    Last edited by MarkFL; December 30th 2011 at 02:15 PM.
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