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Thread: 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 $\displaystyle C(25,5)\cdot C(20,5)$ and got $\displaystyle 823,\!727,\!520.$
    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.


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

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

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


    Included in this long list are these two partitions:

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


    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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    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
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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 $\displaystyle 2m\le n$.

    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:

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

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

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

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

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

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

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

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

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

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

    $\displaystyle N=\frac{1}{p!}\prod_{k=0}^{p-1}{n-km \choose m}$
    Last edited by MarkFL; Dec 30th 2011 at 01:15 PM.
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