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Math Help - Combinatorial problem

  1. #1
    Senior Member Dinkydoe's Avatar
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    Combinatorial problem

    Consider the problem :

    Given S=\left\{1,2,\cdots, 15\right\}. How many subsets S' of 4 different elements can we make, such S' contain no consecutive numbers?

    I like to show that this problem is equivalent with finding the number of solutions of:

    a_1+a_2+a_3+a_4+a_5=14; \ a_1,a_5\geq 0; \ a_2,a_3,a_4\geq 2

    I'm not seeing any connection yet. A little insight please?
    Last edited by Plato; October 16th 2011 at 10:23 AM.
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  2. #2
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    Re: Combinatorial problem

    Quote Originally Posted by Dinkydoe View Post
    I like to show that this problem is equivalent with finding the number of solutions of:

    a_1+a_2+a_3+a_4+a_5=14; \ a_1,a_5\geq 0; \ a_2,a_3,a_4\geq 2
    There are \binom{N+K-1}{N} ways to place N identical objects into K distinct cells.
    Here we have fourteen identical 1's and five distinct variables.
    If we go ahead and put two of the ones into each of a_2,~a_3,~\&~a_4 that leaves eight 1's to place anywhere.
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Re: Combinatorial problem

    That shows that the number we're looking for is {12\choose 8}

    But I don't see how these 2 problems are equivalent. How do they relate..?
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