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Math Help - ways............

  1. #1
    xrajibx
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    ways............

    in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?.............njoy
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    Quote Originally Posted by xrajibx View Post
    in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?.............njoy
    Unless I'm understanding something wrong, in which case tell me:

    Op 1: Choose something for first person: 12 ways
    OP 2: Choose something for second person: 11 ways
    ...
    Op 10: Choose something for tenth person: 3 ways
    Op 11: Choose people for the 2 remaining fruits: 6^2 ways

    Total: 12! * 6^2 / 2 = 18 * 12!

    Other way:
    Op 1: Assign a fruit for each person: 12P10
    Op 2: Choose person for remaining fruit: 6^2
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    Quote Originally Posted by wgunther View Post
    Unless I'm understanding something wrong, in which case tell me:
    I do think that you have misunderstood this problem.
    In your solution, you have assumed that the pieces of fruit are all distinct.
    If that is the case, then the problem is trivial; just count the number of surjections from a set of 12 to a set of 10: Surj(12,10)=6187104000.

    But if we assume that all the apples are virtually identical as well as the oranges are virtually identical, then this becomes an interesting counting problem. Each person gets at least one piece of fruit. It makes a difference is person A gets an apple, an orange, one of each, or two of the same. That is true of each of the other nine people.

    Do you want to rethink your solution?
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    GAMMA Mathematics
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    The answer is at least 10! because 10 pieces of fruit will be split up evenly among the 10 people. Now, there are 2 pieces of fruit left, and order doesn't matter for how these pieces are to be split up among the 10 people. This is simply a combination: 10\choose2 =45, but remember the two pieces of fruit can be two oranges, one apple and one orange, OR two apples. This is four different ways those two pieces of fruit can be split up -- remember the one apple and one orange can switch spots and be a different set.

    10! + 45(4) = 3628980
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    Quote Originally Posted by colby2152 View Post
    The answer is at least 10! because 10 pieces of fruit will be split up evenly among the 10 people.
    Colby2152, how did you get (10!)? Are you assuming 12 different pieces of fruit?
    Did you read my response to the first solution? Assuming 12 different pieces of fruit, makes this a trivial problem.
    The only interesting problem is to assume that the apples are virtually identical as are the oranges.
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    GAMMA Mathematics
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    Quote Originally Posted by Plato View Post
    Colby2152, how did you get (10!)? Are you assuming 12 different pieces of fruit?
    Did you read my response to the first solution? Assuming 12 different pieces of fruit, makes this a trivial problem.
    The only interesting problem is to assume that the apples are virtually identical as are the oranges.
    True, what was I thinking. We have six oranges and six apples. Each person can have an apple or an orange, so that is 2^{10} = 1024 possibilities. The two pieces of fruit that are left can be split up ten different ways if both pieces goto one individual, but the two pieces can take on 3 different forms giving us 30 possibilities. If the pieces of fruit are split up, then there are 10\choose2 *4 = 180 ways the fruit can be split up among two people. There are four ways that a pair of fruits can look like:

    Apple, Apple
    Apple, Orange
    Orange, Apple
    Orange, Orange

    I explained that combination in my last post. Now, let's sum the possibilities:

    1024(30+180)=215040

    EDIT: Somehow, I do not trust this answer considering the first ten pieces of fruit can be evenly split apples and oranges or be a six and four split.
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  7. #7
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    This one of those multiple case problems.
    In one case, someone could get three pieces of fruit: three apples, three oranges, two apples & an orange, or two oranges & an apple. Each of those changes the other distributions.

    Or two people could get two each with a similar set of sub-cases.

    Do you want to see what you can do with that?
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by xrajibx View Post
    in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?.............njoy
    Was just wondering, no idea whether this is right or wrong...
    Say the total number of solutions is S,
    Lets label all the oranges and apples so they are distinct, now the problem reduces to distributing 12 fruits among 10 boxes. So we have 21 \choose 9 solutions. If the total number of solutions is S, we can arrange the apples in 6! ways and similarly for oranges, once we remove labels. So
    S \cdot 6! \cdot 6! = {21 \choose 9}
    So S = \frac{{21\choose 9}}{(6!)^2}

    I hope I am right
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  9. #9
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    Where are you getting those numbers? What is {{21} \choose {9}}?

    Also note \frac {{{21} \choose {9}}} {(6!)^2} = 0.567
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    Lord of certain Rings
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    Quote Originally Posted by Plato View Post
    Where are you getting those numbers? What is {{21} \choose {9}}?

    Also note \frac {{{21} \choose {9}}} {(6!)^2} = 0.567
    Whoops!
    My discounting of solutions was wrong
    I cannot divide by 6!

    Lets try a different approach...

    Can I say the event of distributing apples and oranges are separate?
    I mean I will count the number of ways to distribute apples only and then, for each such distribution I can have all of "oranges distribution".
    I can give 'x1' number of apples to the first guy, x2 to second and so on.
    So I am asking total number of non-negative solutions to
    x1+x2+x3+.....+x10 = 6.
    The total number of solutions for this {10+6-1\choose 6-1}. Similarly for oranges
    So total solutions is ({15\choose 5})^2
    Now, how is it??

    P.S: How do I get a large bracket around that last square number in LaTex??
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  11. #11
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    You overlooked one important fact: Every person gets at least one piece of fruit. That cannot happen when you consider separate cases.
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    Lord of certain Rings
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    Quote Originally Posted by Plato View Post
    You overlooked one important fact: Every person gets at least one piece of fruit. That cannot happen when you consider separate cases.
    Where did he ever say that?? He said any people can get any of those
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  13. #13
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    Quote Originally Posted by Isomorphism View Post
    Where did he ever say that?? He said any people can get any of those
    in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?
    Without that is a trivial problem.
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    Quote Originally Posted by xrajibx View Post
    in how many ways can 6 apples and 6 oranges be distributed among 10 persons provided each get any one of those?.............njoy
    I interpreted that as "provided each person gets either an orange or an apple", so at least one piece of fruit.
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  15. #15
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    Quote Originally Posted by colby2152 View Post
    I interpreted that as "provided each person gets either an orange or an apple", so at least one piece of fruit.
    Now that is correct! But in your solution you assumed that there are 12 different pieces of fruit. If that were the case, it is a trivial problem of counting surjections.
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