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Combinatorics Question

  1. Nov 23, 2012 #1
    So the question was, Let x > 0.

    Prove that x[itex]^{n}[/itex] = [itex]\sum[/itex] [itex]\frac{x!}{x!-k!}[/itex] S(n, k).

    Where the sum goes from k = 1 to n and S(n, k) is th Stirling numbers.

    I believe I have proven what I needed to, but my question is why does x have to be greater than 0? Couldn't we define a function that maps [n] to {-x, ......, 1}.
     
    Last edited: Nov 23, 2012
  2. jcsd
  3. Nov 23, 2012 #2

    Ray Vickson

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    Is this equation accurate? The kth term blows up when x = k! for some k in {1,2,...,n}.

    RGV
     
  4. Nov 23, 2012 #3
    that should be an x! on the bottom... ill fix that
     
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