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Math Help - combinatorics question

  1. #1
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    combinatorics question

    We have N bottles, m red and n blue balls, n > N, m > N.
    We need to know in how many ways is it possible to partition
    all read and blue balls into bottles. There should be at list
    one red and one blue ball in each bottle.
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  2. #2
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    Re: combinatorics question

    Quote Originally Posted by Rafael View Post
    We have N bottles, m red and n blue balls, n > N, m > N.
    We need to know in how many ways is it possible to partition
    all read and blue balls into bottles. There should be at list
    one red and one blue ball in each bottle.
    Two essential questions:
    1) are the bottles identical or distinct?
    2) are the balls identical( except for color) or distinct?
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  3. #3
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    Re: combinatorics question

    Quote Originally Posted by Plato View Post
    Two essential questions:
    1) are the bottles identical or distinct?
    2) are the balls identical( except for color) or distinct?
    All bottles are identical, all red balls are identical, all blue balls are identical.
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  4. #4
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    Re: combinatorics question

    Quote Originally Posted by Rafael View Post
    We have N bottles, m red and n blue balls, n > N, m > N.
    We need to know in how many ways is it possible to partition
    all read and blue balls into bottles. There should be at list
    one red and one blue ball in each bottle.
    Quote Originally Posted by Rafael View Post
    All bottles are identical, all red balls are identical, all blue balls are identical.
    There is an important concept known a integer partitions
    Because "There should be at least one red and one blue ball in each bottle".
    We are partitioning the numbers n-N~\&~m-N
    From that webpage, you can see that the is no simple formula or way of doing this problem.
    Here is an example. Suppose n=15,~m=10,~\&~N=6.
    We think this way. Go ahead and put a red in each bottle leaving 9 reds.
    Go ahead and put a blue in each bottle leaving 4 blues.
    Now there are P(9,6)=26 ways to partition 9 into 6 or fewer summands.
    And P(4,4)=5 ways to partition 4 into 4 or fewer summands.
    The product 26\cdot 5=130 is the number of ways of doing both.
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    Re: combinatorics question

    Thanks a lot
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