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Math Help - Binomial coefficients

  1. #1
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    Binomial coefficients

    <br />
k^k  - {k\choose 1}\left( {k - 1} \right)^k  + {k\choose 2}\left( {k - 2} \right)^k  - {k\choose 3}\left( {k - 3} \right)^k  + \ldots + \left( { - 1} \right)^k {k\choose k}\left( {k - k} \right)^k=?<br />
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    If you are asking

    what that series is equal to i think its \sum_{n=0}^{k}{k\choose n}(k-n)
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  3. #3
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    This what I got in terms of a summation.
    <br />
k^k + \sum_{n=1}^{k}{k\choose n}(k-n)^k(-1)^n<br />
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Yeah

    hes right I did not see the alternating signs
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  5. #5
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    I suppose that the answer should be given in a closed form. I can see that it can be written shorter using the sum symbol ( \sum_{n=0}^{k}{k\choose n}(k-n)^k(-1)^n)...
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  6. #6
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    Quote Originally Posted by james_bond View Post
    <br />
k^k  - {k\choose 1}\left( {k - 1} \right)^k  + {k\choose 2}\left( {k - 2} \right)^k  - {k\choose 3}\left( {k - 3} \right)^k  + \ldots + \left( { - 1} \right)^k {k\choose k}\left( {k - k} \right)^k=?
    That happens to count the number of surjections from a set of k to a set of k objects.
    Or another way it counts surjections from a finite set of k objects onto itself. Those happen to be permutations.
    How ways can we permute k objects. That is your answer.
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