Proving N!/(N-n)! = N^n with Stirling's Approximation | Stat Mech HW

[leright]

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 canceled 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.

 

[stunner5000pt]

what you did with the stirling's approximation is correct

$$\frac{\exp(N\ln(N) - N)}{\exp((N-n)\ln(N) - N + n)}$$

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

o and don't forget N >> n in the end

 

[leright]

what you did with the stirling's approximation is correct

$$\frac{\exp(N\ln(N) - N)}{\exp((N-n)\ln(N) - N + n)}$$

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

o and don't forget N >> n in the end

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.

 

[stunner5000pt]

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.

when you divide you subtract the exponents right?

$$\frac{\exp(N\ln(N) - N)}{\exp((N-n)\ln(N) - N + n)} = \exp\left(N\ln(N) - N)-((N-n)\ln(N) - N + n)\right) = \exp(N\ln(N) - N - (N-n)\ln(N) + N - n)$$

right?

add subtract what you can...

 

[leright]

when you divide you subtract the exponents right?

$$\frac{\exp(N\ln(N) - N)}{\exp((N-n)\ln(N) - N + n)} = \exp\left(N\ln(N) - N)-((N-n)\ln(N) - N + n)\right) = \exp(N\ln(N) - N - (N-n)\ln(N) + N - n)$$

right?

add subtract what you can...

um, of course...that's what I just showed you, except I already canceled the Ns.

 

[leright]

got it. thanks.