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Thread: [COMBINATION]Splitting marbles to boxes + proof

  1. #1
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    [COMBINATION]Splitting marbles to boxes + proof

    Hey guys, can you help me with this HW? John got 30 marbles, 8 white, 8 blue, 5 red and 9 green. He has to split those marbles into 4 boxes with rules. Box no. 1: max 13 marbles, only whites and reds, both of them must be there.
    Box no. 2: max 13 marbles, blues and greens, but the amount of greens and blues must be the same.
    Box no. 3: max 15 marbles, blues and whites, but the amount of blues and whites must be different.
    Box no. 4: max 9 marbles, greens and reds, but only one color at the same time, so just greens or just reds. My task is to solve, how many different ways of splitting them John has. And to make a proof of that. Right now I am really stuck. Thank you really much!
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  2. #2
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Suppose you decide to put red marbles in box #4. Then the only box that can take green marbles in box #2. That means you need 9 green marbles in box #2 and 9 blue marbles in box #2. However, there are only 8 blue marbles, so we cannot have the same number of blue and green marbles in box #2. Therefore, we cannot put red marbles in box #4. This means all of the red marbles are in box #1. Box #1 also needs at least one white marble. So, that leaves the following marbles to place:

    7 white, 8 blue, and 9 green. Box #2 can have 0, 2, 4, 6, 8, 10, or 12 marbles (because they come in pairs). That is six possibilities.
    Suppose box #2 has $2k$ marbles (k blue and k green with $0 \le k \le 6$), then box #4 has $9-k$ green marbles and box #3 has $8-k$ blue marbles. There are 8 ways to distribute the white marbles between boxes 1 and 3, however, one of those ways will yield the same number of blue and white marbles in box 3, so there are 7 possible ways (unless k=0). So, 5 of the arrangements for box #2 yield 7 arrangements of for the white marbles each while 0 marbles in box #2 yields 8 arrangements for the white marbles.

    That means there are $5\cdot 7 + 1\cdot 8 = 43$ different ways to arrange the marbles that satisfy all of the rules.
    Last edited by SlipEternal; Oct 18th 2017 at 12:35 PM.
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Thank you so much. I forgot to mention last thing, if not said opposite, the boxes can be empty. I guess there wonīt be big difference. But I am not sure where to put this condition.
    Last edited by Boglin; Oct 18th 2017 at 01:40 PM.
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Quote Originally Posted by Boglin View Post
    Thank you so much. I forgot to mention last thing, if not said opposite, the boxes can be empty. I guess there wonīt be big difference. But I am not sure where to put this condition.
    Since you did not mention I assumed boxes could be empty. I even stated as much and gave the condition that box 2 had zero marbles in it (empty).
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    I have got one question. You said: Box #2 can have 0, 2, 4, 6, 8, 10, or 12 marbles (because they come in pairs). That is six possibilities.
    But thatīs 7 possibilities or am I missing something?
    Thank you
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Quote Originally Posted by Boglin View Post
    I have got one question. You said: Box #2 can have 0, 2, 4, 6, 8, 10, or 12 marbles (because they come in pairs). That is six possibilities. But thatīs 7 possibilities or am I missing something?
    Yes, I think that you are missing the point that $0$ does not give a pair.
    That is there are only six numbers of marbles: $2,~ 4,~ 6,~ 8,~ 10, \text{ or } 12$
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    I have similar problem.
    Peter has 5 white, 6 red, 10 green and 7 blue marbles.
    4 boxes
    a# max 13 marbles,only WHITE and RED, but same amount of them
    b# max 11 marbles, only GREEN and BLUE, can't be empty
    c# max 11 marbles, only BLUE and WHITE, but only white or blue ones, not both colours same time
    d# max 10 marbles, only RED and GREEN, but different amount of them.
    0 marbles in box is also viable choice.
    I would be very glad if you could help me, because I'm lost .. really appreciate all your help and time
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Quote Originally Posted by Boglin View Post
    I have got one question. You said: Box #2 can have 0, 2, 4, 6, 8, 10, or 12 marbles (because they come in pairs). That is six possibilities.
    But thatīs 7 possibilities or am I missing something?
    Thank you
    You are correct. My mistake
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Does it change the result? Thank you
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  10. #10
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    Re: [COMBINATION]Splitting marbles to boxes + proof

    Quote Originally Posted by Boglin View Post
    Does it change the result? Thank you
    No
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