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StatMech: Binomial approximation
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[QUOTE="rsaad, post: 4511603, member: 442750"] I looked at the stirling formula derivation but I don't know how it is helpful here. So I have solved it the other way. [N-(n-(n-0))] [N-(n-(n-1))] [N-(n-(n-2))] [N-(n-(n-3))]...[N-( n-(3) )][N-( n-(2) )][N-( n-(1) )] For N>>n, using this approximation once, I get n terms: [N-n] [N-n] [N-n] ... [N-n] [N-n] [N-n] using the approximation again, [N] [N] [N] ...[N] [N] [N] =N^n My question is can I use this approximation selectively like I did in the above two steps. Secondly, how was sterling formula derivation helpful? I used the x! formula and I get exponential(-n). Because n<<N, this term is big, making the entire answer zero. [/QUOTE]
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StatMech: Binomial approximation
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