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Stat mech hw.

  1. Nov 14, 2006 #1
    I am supposed to show that N!/(N-n)! = N^n where 1<<n<<N

    I used stirling's approximation to show that N! = e^(NlnN-N) and (N-n)! = e^[(N-n)ln(N-n) - N + n].

    I took the ratio of these two terms to get e^[NlnN-N-(N-n)ln(N-n) + N - n]. I cancelled terms and get N!/(N-n)! = N^N/[(N-n)^(N-n)e^n], which isn't N^n.

    btw, stirlings says that lnN! = NlnN - N.

    Can someone give me a hint? That would be great.
  2. jcsd
  3. Nov 14, 2006 #2
    what you did with teh stirling's approximation is correct

    [tex] \frac{\exp(N\ln(N) - N)}{\exp((N-n)\ln(N) - N + n)} [/tex]

    so when you divide what do you do with exponents??

    o and dont forget N >> n in the end

  4. Nov 14, 2006 #3
    right, I got that far, and ended up with e^(NlnN-(N-n)ln(N-n)-n) = N^N/[(N-n)^(N-n)e^n].

    If I say that N-n is about equal to N I get e^(-n), which is obviously wrong.
  5. Nov 14, 2006 #4
    when you divide you subtract the exponents right?

    [tex] \frac{\exp(N\ln(N) - N)}{\exp((N-n)\ln(N) - N + n)} = \exp\left(N\ln(N) - N)-((N-n)\ln(N) - N + n)\right) = \exp(N\ln(N) - N - (N-n)\ln(N) + N - n) [/tex]


    add subtract what you can...
  6. Nov 14, 2006 #5
    um, of course...that's what I just showed you, except I already cancelled the Ns.
  7. Nov 14, 2006 #6
    got it. thanks.
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