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Homework Help: Basketball roster probability problem

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data
    A basketball coach has 10 boys and 10 girls on his roster. In how many ways can the coach partition the group of 20 into 4 teams of 5 students such that there are at least 2 boys and 2 girls on each team ?


    3. The attempt at a solution

    I tried nCr(10,3)*nCr(10,2) + nCr(7,3)*nCr(8,2) + nCr(6,3)*nCr(4,2)=6500

    I think it's correct but Im not sure.

    Thanks.
     
  2. jcsd
  3. Feb 19, 2010 #2

    vela

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    Re: Probability.

    I think that number's way too small. The number of ways you can form just one team is already [C(10,3)*C(10,2)]*2=5400*2=10800. (The factor of 2 is there for this example because you can have 3 boys and 2 girls or 2 boys and 3 girls.)

    Try working out what you get if you had 3 boys and 3 girls and had to form three teams with at least one boy and one girl each. It's easy enough to write down all the combinations and see how to count them.
     
  4. Feb 19, 2010 #3
    Re: Probability.

    hey,
    thanks for the response.
    So, basically I have to multiply everything by 2, correct?
    2*[C(10,3)*C(10,2)+C(7,3)*C(8,2)+C(6,3)*C(4,2)]=13000

    ??
     
  5. Feb 19, 2010 #4

    vela

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    Re: Probability.

    No, that factor of two is only there because forming only one team leaves students that weren't chosen. When you form four teams, every student gets assigned to a team. If you count the ways to form all four teams, you end up covering every possible combination, so there's no need to double the number.

    For instance, suppose you had three girls (A, B, and C) and three boys (0, 1, and 2), and you want to form two teams, each with at least one boy and one girl. One possible team is AB0, so the other team has to be C12. When you count AB0, you're already counting C12. If you were to count the team C12 again separately, you'd be double-counting.
     
  6. Feb 19, 2010 #5
    Re: Probability.

    I understand what you are saying, but if I don't need to multiply by two, I don't understand my mistake.

    I understand that I don't need to multiply by two because every kid gets assigned.

    As far as I understand it,

    2*(C(10,3)*C(10,2))
    here we multiply by two because we can have either a team of three boys or a team of three girls, right?

    now for the second one we also can have either, so multiply the second one by two also ?
    However, for the last two there is 1 option only.
    So, we multiply by two the first two ?

    Is this what you mean ?

    OR

    is my combination incorrect ?

    Thanks.
     
  7. Feb 19, 2010 #6

    vela

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    Re: Probability.

    Your calculating the ways you can form the first team, the second team, and the third team correctly. You're just not calculating the total number of possibilities correctly.

    Think about flipping 3 coins. Your calculation is akin to saying since the each coin can come up heads or tails, there are 2+2+2=6 ways possible outcomes of flipping 3 coins, but there's actually 8: hhh, hht, hth, htt, thh, tht, tth, and ttt.
     
  8. Feb 19, 2010 #7
    Re: Probability.

    Well,
    if this is incorrect:
    2*(C(10,3)*C(10,2)) + 2*(C(7,3)*C(8,2)) + C(6,3)*C(4,2)

    is incorrect, I don't know how to get the correct answer.
    Can you please tell me how to get the correct answer?

    Thanks.

    EDIT:

    so basically I need to multiply them ?
    if we go according to the possibilities of filliping a coin, we need to multiply them all ?
    so what we get is :

    4*(C(10,3)*C(10,2))*(C(7,3)*C(8,2)) *(C(6,3)*C(4,2))

    ?
     
  9. Feb 19, 2010 #8
    Re: Probability.

    In your edit you should be dividing by 4 rather than multiplying because the two boy teams (teams with 3 boys) can be switched and it wouldn't make a difference to the partition, same goes with the two girl teams (3 girls on the team).
     
  10. Feb 20, 2010 #9
    Re: Probability.

    I see.
    THanks.
     
    Last edited: Feb 20, 2010
  11. Feb 20, 2010 #10
    Re: Probability.


    Now that I am reading it again, why did we need just to divide by 4 and not also multiply by ?

    I think Vela's method was correcct
    There are 2 options how we can form the first group 2*[C(10,3)*C(10,2)]- it can either be 3B 2G OR 3G 2B. therefore, we need to multiply by two ....
    I guess we need to multiply the numerator by 8 ?! and divide by 4 ? (I also understand why you are saying that we need to divide and I think you are right about this. But I am not sure why not to multiply ?!)

    THanks.
     
  12. Feb 20, 2010 #11
    Re: Probability.

    That reasoning is correct for the first group, but try to continue it for the 2nd and 3rd groups (particularly the 3rd), and you'll begin to see where the difficulty lies in that approach (you won't end up with the same product as you wrote).

    Giving an interpretation to the product you wrote down, you can think of the process as selecting a boy group, then another boy group, then a girl group, then another girl group (here boy group means group containing 3 boys). How do you select the first boy group? C(10,3) * C(10,2). Once this group is selected how do you select the second boy group? C(7,3) * C(8,2). Then the first girl group C(4,2) * C(6,3) and finally the last girl group C(2,2) * C(3,3). Once you do this you've over counted by a factor of 4, since you've counted the boy groups A then B, and B then A as different. Same goes for the girl groups.
     
  13. Feb 20, 2010 #12

    vela

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    Re: Probability.

    Remember, you're counting possible partitions, not possible teams. As Mandark noted, you can think of forming two boy teams and two girl teams. Every partition will contain a pair of 3B-2G teams and pair of 3G-2B teams. If you count the 3B-2G teams separately from the 3G-2B teams, you will be counting each partition that contains a specific combination of those teams multiple times.

    Again, to make what we're saying more concrete, I encourage you to work out what happens if you have fewer boys and girls and form smaller teams. You can list every possible team and possible partitioning and count them. Then try to calculate those numbers mathematically, and you'll hopefully see how the calculations work out.
     
  14. Feb 20, 2010 #13
    I don't get it... Is it totally wrong ?
    I understand what u r saying, but how do u know which team is first ? We have 4 factorial combinations ( ggbb,gbgb,bbgg etc.)

    thanks.
     
  15. Feb 20, 2010 #14
    I'm getting more confused, what do you mean calculating them seperately is not gonna work? I'm not calculating them seperately.
    Can you tell me what the answer is so I could compare it with my answer and see my mistake? Tnx
     
  16. Feb 20, 2010 #15
    Re: Probability.

    Actually, after reading your posts again, I think I understand why I need to divide and not multiply by 4.
    It doesn't matter if we start with the boy or the girl team, it get completed by the whole combination.

    Thank you very much for the help.
     
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