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Thread: Combinatorics with several conditions

  1. #1
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    Unhappy Combinatorics with several conditions

    Hello. I have a math problem that I need to solve, however Im really stuck and also kind of depressed now.

    We have 30 balls. 6 red, 7 white, 9 green and 8 blue. We want to seperate them to 4 different boxes.
    ∑ First box can only contain 13 or less balls, which can be either red or white. Number of red and white balls in this box must be equal.
    ∑ Second box can contain 13 or less balls, which can be either green or blue. There must be at least one ball from each of these colors.
    ∑ Third box can contain 10 or less balls, which can be either blue or red. However there can only be be balls from one of these colors. Not both.
    ∑ Fourth box can contain 8 or less balls, which can be either white or green. This box canít stay empty.

    If the condition doesnít stay otherwise, boxes can remain empty as well.
    Im supposed to determine the number of possible solutions AND
    prove that my solution is correct. Unfortunately, creating a tree or
    creating a computer program is not allowed. It's supposed to be as simple and exact as possible. I think the correct way is to seperate conditions to smaller parts
    (balls go to the box number x or y) and then combine them using
    intersection and union. Maybe use inclusion and exclusion principle.
    Do you see it as a viable option? Can you please show me how? I
    spent too much time on this already to just let it pass
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  2. #2
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    Re: Combinatorics with several conditions

    Quote Originally Posted by Ormis View Post
    Hello. I have a math problem that I need to solve, however Im really stuck and also kind of depressed now.

    We have 30 balls. 6 red, 7 white, 9 green and 8 blue. We want to seperate them to 4 different boxes.
    ∑ First box can only contain 13 or less balls, which can be either red or white. Number of red and white balls in this box must be equal.
    ∑ Second box can contain 13 or less balls, which can be either green or blue. There must be at least one ball from each of these colors.
    ∑ Third box can contain 10 or less balls, which can be either blue or red. However there can only be be balls from one of these colors. Not both.
    ∑ Fourth box can contain 8 or less balls, which can be either white or green. This box canít stay empty.

    If the condition doesnít stay otherwise, boxes can remain empty as well.
    Im supposed to determine the number of possible solutions AND
    prove that my solution is correct. Unfortunately, creating a tree or
    creating a computer program is not allowed. It's supposed to be as simple and exact as possible. I think the correct way is to seperate conditions to smaller parts
    (balls go to the box number x or y) and then combine them using
    intersection and union. Maybe use inclusion and exclusion principle.
    Do you see it as a viable option? Can you please show me how? I
    spent too much time on this already to just let it pass
    See here This question has been asked and answered.
    Thanks from topsquark
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