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Math Help - combinatorics question

  1. #1
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    combinatorics question

    Hello everybody,
    This is probably a simple combinatorics question.

    - We do have A red balls, B green balls and C yellow balls ( A, B, C is their count, so they are totall A+B+C in count). If we combine them together into n-pairs how many are their unique combinations?

    I tried for the combination into 2-pairs and i solved it for special cases (example A=3, B=1, C=1).
    \left(\begin{array}{cc}3\\2\end{array}\right)+<br />
\left(\begin{array}{cc}A-1\\2\end{array}\right)+<br />
\left(\begin{array}{cc}B-1\\2\end{array}\right)+<br />
\left(\begin{array}{cc}C-1\\2\end{array}\right)=4
    However if i try to generalize it, I fail. I know that I am confused and this is probably wrong proccess to the problem.

    Thank you all for your time.
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  2. #2
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    Quote Originally Posted by gdmath View Post
    We do have A red balls, B green balls and C yellow balls ( A, B, C is their count, so they are totall A+B+C in count). If we combine them together into n-pairs how many are their unique combinations?
    That is a somewhat vague description.
    By 'pairs' do you mean ordered pairs, simple sets of two each, or something else altogether?
    It is simply not clear what a "3-pair" could be.
    Can you clear this up? Give an example: say A=3,~B=2,~C=1.
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  3. #3
    Senior Member Sambit's Avatar
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    I suppose the question asks for non-ordered pairs and balls in each colour are non-identical.

    RED-RED pairs: {{A}\choose{2}}

    GREEN-GREEN pairs: {{B}\choose{2}}

    YELLOW-YELLOW pairs: {{C}\choose{2}}

    RED-GREEN pairs: {{A}\choose{1}}{{B}\choose{1}}

    GREEN-YELLOW pairs: {{B}\choose{1}}{{C}\choose{1}}

    YELLOW-RED pairs: {{C}\choose{1}}{{A}\choose{1}}

    Add these to get the total number of unique pairs. This is actually the total number of pairs each of which are different.


    Quote Originally Posted by gdmath View Post
    I tried for the combination into 2-pairs and i solved it for special cases (example A=3, B=1, C=1).
    \left(\begin{array}{cc}3\\2\end{array}\right)+<br />
\left(\begin{array}{cc}A-1\\2\end{array}\right)+<br />
\left(\begin{array}{cc}B-1\\2\end{array}\right)+<br />
\left(\begin{array}{cc}C-1\\2\end{array}\right)=4
    If B,C are 1, the above expression becomes meaningless.

    EDIT: By the way, after working it out it seems to me here PAIR means any number of balls taken together; is that right?
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  4. #4
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    First of all thank you for your time and your response.

    Let me give you an example:

    We have:
    - 4 Balls of red (A=4)
    - 3 Balls of green (B=3)
    - 5 Balls of yellow (C=5)

    If we pick up 3 (n=3) balls each time, how many possible combinations (unique combinations) can have?
    We do not care about the distribution of combinations. We only care if we can count the unique combinations.

    for this example, by observation we do have 10 unique combinations:
    (we do not care for the order)

    Red - Red - Green
    Red - Green - Green
    Yellow - Green - Green
    Yellow - Yellow - Green
    Red - Red - Yellow
    Red - Yellow - Yellow
    Red- Green- Yellow
    Red-Red-Red
    Green-Green-Green
    Yellow - Yellow-Yellow



    Is it possible to use a generalized formula with A, B, C and (mostly) n?

    THANK YOU A LOT
    Last edited by gdmath; April 10th 2011 at 09:33 AM.
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  5. #5
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    Quote Originally Posted by gdmath View Post
    We have:
    - 4 Balls of red (A=4)
    - 3 Balls of green (B=3)
    - 5 Balls of yellow (C=5)
    If we pick up 3 (n=3) balls each time, how many possible combinations (unique combinations) can have?
    We do not care about the distribution of combinations. We only care if we can count the unique combinations. for this example, by observation we do have 7 unique combinations:
    (we do not care for the order)
    Thank you for the nice clarification.
    In this particular example, the answer is simple because the number of selections, n=3, is not more than the minimum of color calls, the greens.
    The answer is \dbinom{3+3-1}{3}=10.
    In your count of seven your miss three selections.

    In the case n>\min\{A,B,C\} the process is more complicated.
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  6. #6
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    In the case n>\min\{A,B,C\} the process is more complicated.
    Let me assume that there is no a generalized formula of doing that.
    Furthermore the process depends on the arithetic data and might change on specific ocussions?

    THANK YOU A LOT.
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  7. #7
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    Quote Originally Posted by gdmath View Post
    Let me assume that there is no a generalized formula of doing that. Furthermore the process depends on the arithetic data and might change on specific ocussions?
    Well if n\le\min\{A,B,C\} then there general formula:
    \dbinom{n+2}{n}.
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