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Thread: [SOLVED] distributing 'N' pennies to "K" children.

  1. #1
    baz
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    [SOLVED] distributing 'N' pennies to "K" children.

    Can anybody through a light on this? I am not able toget the answer.

    In how many ways can we distribute "N" coins to 'K" children so that everyone gets atleast one.

    thnx in advance.

    should it be n!/k! ?
    if not, why?
    Last edited by mr fantastic; Apr 2nd 2010 at 05:16 AM. Reason: Merged posts
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    Quote Originally Posted by baz View Post
    Can anybody through a light on this? I am not able toget the answer.
    In how many ways can we distribute "N" coins to 'K" children so that everyone gets atleast one.
    We assume that the coins are identical.
    The number of ways to place N identical objects into K distinct cells is \binom{N+K-1}{N}.
    But the number of ways to do it so that no cell is empty equals
    \binom{(N-K)+(K-1)}{N-K}=\binom{N-1}{N-K} .
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  3. #3
    baz
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    Quote Originally Posted by Plato View Post
    We assume that the coins are identical.
    The number of ways to place N identical objects into K distinct cells is \binom{N+K-1}{N}.
    But the number of ways to do it so that no cell is empty equals
    \binom{(N-K)+(K-1)}{N-K}=\binom{N-1}{N-K} .
    I dont get it that how the number of ways to place N identical objects in K distinct cells is \binom{N+K-1}{N}?

    Can you explain it please?
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    Quote Originally Posted by baz View Post
    I dont get it that how the number of ways to place N identical objects in K distinct cells is \binom{N+K-1}{N}?
    Can you explain it please?
    This is not a tutorial service.
    You are expected to do your own research.
    You should do a web search for such topics: Multi-sets or multi-selections.
    Scroll down to find the discussion on multiset coefficients.
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