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Math Help - Combinations

  1. #1
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    Combinations

    Ballots for municipal elections usually list candidates for several different positions. If a resident can vote for a mayor, two councillors, a school trustee, and a hydro commissioner, how many combinations of positions could the resident choose to mark on one ballot?

    (2x3x2x2)-1
    =24-1
    =23?

    The answer in in the book says 31 but i don't understand how or if the book is wrong.
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  2. #2
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    Quote Originally Posted by KobeBryant
    Ballots for municipal elections usually list candidates for several different positions. If a resident can vote for a mayor, two councillors, a school trustee, and a hydro commissioner, how many combinations of positions could the resident choose to mark on one ballot?

    (2x3x2x2)-1
    =24-1
    =23?

    The answer in in the book says 31 but i don't understand how or if the book is wrong.
    I think you are not saying the whole story. For example, two councillors out of how many?
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    That's the full question copied word for word, I'm thinking that the back of the book is wrong.
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  4. #4
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    It seems the question is asking for the different ways the voter could vote or not vote for the 5 different positions. For example, the voter could vote for only mayor and hydro commissioner. That's 2 positions. There are \binom{5}{2} = 10 ways to vote for 2 positions. The total number of ways to vote is

    \sum_{k=1}^5 \binom{5}{k} = 5 + 10 + 10 + 5 + 1 = 31.
    Last edited by JakeD; May 1st 2006 at 08:04 PM.
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  5. #5
    Super Member Rebesques's Avatar
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    Just wanted to remind you a useful formula:

    <br />
\sum_{k=0}^n \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}1^k1^{n-k}= (1+1)^n=2^n. <br />

    So <br />
\sum_{k=1}^5 \binom{5}{k} = 2^5-1 = 31.<br />
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  6. #6
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    Quote Originally Posted by Rebesques
    Just wanted to remind you a useful formula:

    <br />
\sum_{k=0}^n \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}1^k1^{n-k}= (1+1)^n=2^n. <br />
    I did not know this. Thanks for pointing it out.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques
    Just wanted to remind you a useful formula:

    <br />
\sum_{k=0}^n \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}1^k1^{n-k}= (1+1)^n=2^n. <br />
    I like that one!

    -Dan
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