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Math Help - Simple question about propabilites

  1. #1
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    Simple question about propabilites

    Hi everybody,
    Combinatorics always confusing me. I ask you to give me the type of a classic issue in propability.

    We have N boxes. Each box contains P0...Ps elements (so means s to count).

    If we pick up an element from each box:
    - How many will be the UNIQUE comhinations of the elements?
    if you provide and an analytic solution, it would be greatful.
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  2. #2
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    Quote Originally Posted by gdmath View Post
    We have N boxes. Each box contains P0...Ps elements (so means s to count).
    If we pick up an element from each box:
    - How many will be the UNIQUE comhinations of the elements?
    if you provide and an analytic solution, it would be greatful.
    Not a good description of the problem.
    Are the elements in the same box identical?
    If not how many of each kind?

    Can two boxes have any common elements?
    If so, how many?

    I really think that you should try to rewrite the to make all the condition on the boxes and on the elements as clear as possible.
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  3. #3
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    Ok,

    We have N boxes.

    Assume each box contains elements that can be described as a number.

    So each box contains numbers from 1 to s (1,2,3,...,s). So these elements are deferent from each other.

    Next I say "each box". This mean that the first box has exactly the same elements with any other box.

    So if we pick one element from each box (total N elements) how many unique combinations will have?

    (Unique means that if i have the combination 123456...N i cannot accept the combination 213456..N)

    Hope that i am clear.

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  4. #4
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    MUCH better.
    Letís give an example where there are sill some questions.
    Suppose we have six boxes each looks this: \left\{ {0,1,2,3,4,5,6} \right\}.
    Choosing one element from each box we could have this sample: \left\{ {0,1,3,4,5,6} \right\}. Clearly six different elements.

    But we could also have \left\{ {0,1,1,4,0,0} \right\} with some repetitions.
    Now as I understand you mean this is the same as \left\{ {0,0,0,1,1,4} \right\}.
    Is that correct?

    Under this understanding the answer is \binom{6+7-1}{6}.
    That is the number of ways to select six items from a variety of seven.

    \binom{N+S-1}{N} is the number of ways to select N items from a variety of S.

    If this is not what you mean, letís try again.
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  5. #5
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    Yes exactly this i need.

    I knew it was something with binomial, but not exactly what...

    First of all thank you

    Now look something interesting,

    The way i was trying to realize the problem,

    when there are two boxes the probabilty is (s elements in each box):

    \sum_{i=1}^{s+1} i = \frac{1}{2} s(s+1)=\binom{s+2-1}{2}

    When we have three Boxes:
    \sum_{j=1}^{j+2} ( \sum_{i=1}^{j+1} i ) = \frac{1}{6} s(s+1)(s+2)=\binom{s+3-1}{3}

    And goes on recursively at sums...

    Thank you again
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