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Math Help - Multiset

  1. #1
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    Multiset

    Figured out, can this be deleted?
    Last edited by Jrb599; January 28th 2008 at 10:52 AM.
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  2. #2
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    \sum\limits_{k = 0}^n {{{k + 1} \choose k}\left( {n + 1 - k} \right)}
    Can you explain this?
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  3. #3
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    Plato, that is the answer, but I have no idea how to get that.
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  4. #4
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    ^^actually unless I didn't something wrong, that doesn't hold for n=2
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  5. #5
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    The actual sum is this <br />
\sum\limits_{k = 0}^n {\left( {k + 1} \right)\left( {\begin{array}{*{20}c}<br />
   {n + 1}  \\<br />
   {n - k}  \\<br />
\end{array}} \right)} . You can choose any number form 0 to n letters.

    Say n=10; then you may choose 3 a’s and a b along with any six numbers from 1 to 7. In this case, you have chosen 4 letters and 6 numbers. There are 5 ways to pick 4 letters: aaaa, aaab, aabb, abbb, & bbbb. There are 7 ways to pick six numbers from 1 to 7.

    Now we sum on all the cases.
    Last edited by Plato; January 27th 2008 at 01:21 PM.
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  6. #6
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    Plato then n = 2


    should give

    (a,a) (a,b) (a,1) (a,2) (a,3)
    (b,b) (b,1) (b,2) (b,3)
    (1,2) (1,3)
    (2,3)

    Which is twelve 2 - combinations for n =2

    but the formula you put says there should be 10.
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  7. #7
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    Quote Originally Posted by Jrb599 View Post
    Plato then n = 2
    Which is twelve 2 - combinations for n =2
    but the formula you put says there should be 10.
    Yes you are correct it should be
    <br />
\sum\limits_{k = 0}^n {\left( {k + 1} \right)\left( {\begin{array}{*{20}c}<br />
   {n + 1}  \\<br />
   {n - k}  \\<br />
\end{array}} \right)}
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  8. #8
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    Plato, sorry to bug you with questions

    but how did you start your logic process? Why do you multiple the two terms?

    Also how did you know to sum them up?
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