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Thread: Combinations question

  1. #1
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    Question Combinations question

    Combinations question-emek.png
    For part (a)

    Would i be right to say:

    There's sequence of 30 0's and 9 1's.

    C(k+n-1,k) = (k+n-1)!/k!(n-1)! =

    C(39,30) = 39!/30!9!

    The solution someone else gave me is different to my assumptions.

    Please help

    Solution someone else told me:
    Combinations question-ddsd.png
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  2. #2
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    Re: Combinations question

    Quote Originally Posted by inayat View Post
    Click image for larger version. 

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    For part (a)
    Would i be right to say:
    There's sequence of 30 0's and 9 1's.
    C(k+n-1,k) = (k+n-1)!/k!(n-1)! or C(39,30) = 39!/30!9! YES!
    b) There are five odd number boxes. If we go ahead and put one ball in to each of those five then distribute the remaining 25 balls into ten boxes in $\dbinom{25+10-1}{25}$ ways. Then odd boxes are nonempty.

    c) This can be a trick thought experiment. Zero is an even number, so an empty box contains an even number of balls. Now think if "gluing" the balls into pairs of two each. Now put these 15 identical objects into the ten boxes.
    That can be done in $\dbinom{15+10-1}{15}$ ways. Each box contains an number of the original balls.
    Thanks from inayat
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  3. #3
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    Re: Combinations question

    Quote Originally Posted by Plato View Post
    b) There are five odd number boxes. If we go ahead and put one ball in to each of those five then distribute the remaining 25 balls into ten boxes in $\dbinom{25+10-1}{25}$ ways. Then odd boxes are nonempty.

    c) This can be a trick thought experiment. Zero is an even number, so an empty box contains an even number of balls. Now think if "gluing" the balls into pairs of two each. Now put these 15 identical objects into the ten boxes.
    That can be done in $\dbinom{15+10-1}{15}$ ways. Each box contains an number of the original balls.
    Thanks that helps
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