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Math Help - Permutation and Combination

  1. #1
    Senior Member pankaj's Avatar
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    Permutation and Combination

    There are 5 mangoes,10 apples and 15 oranges.Find number of ways to distribute 15 fruits to two persons.
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    Quote Originally Posted by pankaj View Post
    There are 5 mangoes,10 apples and 15 oranges.Find number of ways to distribute 15 fruits to two persons.
    There are \binom{X+2-1}{2-1}=(X+1) ways to give X identical items to two people.
    So what is the solution to your question?
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  3. #3
    Super Member malaygoel's Avatar
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    There are 30 fruits

    And does two mangoes are different from each other...or they can be assumed identical?
    e.g. first mango is same as second mango?????
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  4. #4
    Super Member Random Variable's Avatar
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    How about expanding  \Big((1+x+x^{2}+...+x^{5})(1+x+x^2+...+ x^{10})(1+x+x^2+...+x^{15})\Big)^{2} and finding the coefficient of  x^{15} ?
    Last edited by Random Variable; July 7th 2009 at 09:58 AM.
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  5. #5
    Super Member Random Variable's Avatar
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    11304?
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  6. #6
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Plato View Post
    There are \binom{X+2-1}{2-1}=(X+1) ways to give X identical items to two people.
    So what is the solution to your question?
    But before that you need to select 15 fruits.And then when you distribute them all fruits are not identical i.e there are 3 different varieties of fruits.Your formula works only when all objects are alike and of same genre.

    Quote Originally Posted by Random Variable View Post
    How about expanding  \Big((1+x+x^{2}+...+x^{5})(1+x+x^2+...+ x^{10})(1+x+x^2+...+x^{15})\Big)^{2} and finding the coefficient of  x^{15} ?
    This would be number of ways to select 15 fruits from given 30 fruits and comes out to be 66.
    Now,what is to be done.

    Unless we take into account each selection separately then only can the answer be determined.
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  7. #7
    Super Member Random Variable's Avatar
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    This would be number of ways to select fruits from given fruits and comes out to be .
    It's not 66. I squared the entire quantity. If there were n people, I would have rasied it to the nth power.
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  8. #8
    Senior Member pankaj's Avatar
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    15 fruits from available 30 fruits are to be distributed among 2 persons. It is not that one person gets one set of 15 fruits and the other gets a different set of 15 fruits.

    For malaygoel

    Fruits of the same type are identical
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  9. #9
    Super Member Random Variable's Avatar
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    nevermind
    Last edited by Random Variable; July 7th 2009 at 08:48 PM.
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  10. #10
    Member billym's Avatar
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    Would it be

    [30!/(5!10!15!)]*C(30,15)*C(16,15)

    Like... the number of permutations of the 30 fruits mulitplied by the number of combinations of size 15 from 30 possible fruits mulitiplied by the number of different ways you can distribute 15 fruits to 2 people...

    Way off?
    Last edited by billym; July 7th 2009 at 07:47 PM.
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  11. #11
    Member billym's Avatar
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    oops
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  12. #12
    Super Member malaygoel's Avatar
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    Quote Originally Posted by pankaj View Post
    There are 5 mangoes,10 apples and 15 oranges.Find number of ways to distribute 15 fruits to two persons.
    let us pick a collection of 15 fruits:
    2 mangoes, 7 apples and 6 oranges.

    now, these can be divided among two people in 3*8*7=168 ways
    (these include the cases in which one of them don't get anything).

    Let us say that this case corresponds to (3x^2)(8x^7)(7x^6)

    Then all the solution will be given by \sum [(n+1)x^n][(m+1)x^m][(16-n-m)x^{15-n-m}]...subject to constraints:

    0 \leq n \leq 5

    0 \leq m \leq 10

    If the reasoning till now is true, then the solution of given problem is the coefficient of x^{15} in
    (1+2x+...+5x^4+6x^5)(1+2x+...+10x^9+11x^{10})(1+2x  +...+15x^{14}+16x^{15})
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  13. #13
    Super Member Random Variable's Avatar
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    Yeah, my approach would only work if all the fruits were the same.
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