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Math Help - combinatorial voting problem

  1. #1
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    combinatorial voting problem

    1) how many ways can you vote if there are 4 candidates you get 12 votes but you are not required to use all 12.

    x1+x2+x3+x4+x5 = 12
    C(12+5,12)= 6,188
    is this correct?

    2) how many different dinners can be created if
    4 choices of pasta(only can choose 1), 5 choices of sauces(only can choose 1),2 choices of salads(only can choose 1), 4 choices of meats(can choose as many as you want assume they are different ie meat meat isnt allowed), 5 additional toppings(choose up to 3)
    Last edited by canyiah; March 22nd 2010 at 04:14 PM.
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  2. #2
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    Combination with repetition

    Hello canyiah
    Quote Originally Posted by canyiah View Post
    1) how many ways can you vote if there are 4 candidates you get 12 votes but you are not required to use all 12.

    x1+x2+x3+x4+x5 = 12
    C(12+5,12)= 6,188
    is this correct?
    I think this should be \binom{16}{12} = 1820.

    The problem is equivalent to choosing 12 items from 5, with repetition being allowed. The 12 items are the 12 votes, each for a particular candidate or none; the 5 are the 4 candidates plus a 'null' vote, where the vote is not cast for any of the candidates.


    Now the number of ways of choosing k items from n, with repetition allowed, is:
    \binom{n+k-1}{k} (See here, for example.)
    So when k = 12 and n = 5, this is:
    \binom{16}{12} = 1820
    2) how many different dinners can be created if
    4 choices of pasta(only can choose 1), 5 choices of sauces(only can choose 1),2 choices of salads(only can choose 1), 4 choices of meats(can choose as many as you want assume they are different ie meat meat isnt allowed), 5 additional toppings(choose up to 3)
    Since repetition is not allowed in the choice of meat, I'm assuming that it is not allowed in the choice of additional topping either. I'm also assuming that you must choose exactly 1 pasta, 1 sauce and 1 salad, but that you can choose 0 meats and 0 additional toppings (since the phrases 'as many as you want' and 'up to' are used for these items).

    Then we have:
    Number of choices of pasta: 4

    Number of choices of sauce: 5

    Number of choices of salad: 2

    Number of choices of meat: 2^4=16, since each of the 4 meats may be either chosen or rejected

    Number of choices of additional topping: \binom50+\binom51+\binom52+\binom53=26
    So, if my assumptions are correct, I reckon the number of different dinners is:
    4\times5\times2\times16\times26=16640
    Grandad
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