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Math Help - Multinomial coefficient question

  1. #1
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    Multinomial coefficient question

    Hi everyone! I don't quite understand what is the difference between these 2 questions and I was hoping if someone could explain it to me:

    1. Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible? ANS: 10! / (5!5!)

    I understand this solution because it's just following the formula.

    2. In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible? ANS: [10! / (5!5!)] / 2!

    I don't understand what the author means when he says that 'now the order of the two teams is irrelevant'. I still don't quite understand why I have to half the probability and what is the difference between this question and the one before. Can someone help me please?

    Thanks in advance! ^^
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  2. #2
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    On the first case each team will play in a different league. If we number the kids from 0 to 9:

    Team A: 0-1-2-3-4 Team B: 5-6-7-8-9 (situation 1)

    is different from

    Team A: 5-6-7-8-9 Team B: 0-1-2-3-4 (situation 2)

    Because the teams will play in different leagues. Imagine you were the kid 9. It would matter to you whether you play on Team A that goes to Division1 or Team B that goes to Division2.

    On the second question you have kids on a playground ready to play a basketball game, situation 1 and situation 2 are exactly the same. The composition of the teams is equal on both situations and for a kid it's equal to play on Team A or Team B.
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  3. #3
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    Quote Originally Posted by deathstarx View Post
    Hi everyone! I don't quite understand what is the difference between these 2 questions and I was hoping if someone could explain it to me:

    1. Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible? ANS: 10! / (5!5!)

    I understand this solution because it's just following the formula.

    2. In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible? ANS: [10! / (5!5!)] / 2!
    This is this simple. In the first example, it makes a difference to me if I were chosen to be on team B. I would want to be on the A-team.

    Under the second scenario, it is only group membership that matters.
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  4. #4
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    Thanks! I think I get it then.
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