Any ideas on finding a good estimate/approximation for $\frac AB$ where $A = N^L$ and $B = {N+L\choose N}$?
$N^L$ vs. ${N+L\choose N}$
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combinatorics
binomial-coefficients
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2You could try applying Stirling on the factorials implicit in the binomial coefficient... – 2011-05-02
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1I don't understand the notations $N^L$ and $C_{N+L}^N$. What do those mean? – 2011-05-02
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0@Mitch: I am taking $N^L$ as the exponential and $C_{N+L}^N$ as the binomial coefficient of $N+L$ choose $N$ – 2011-05-02
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1In what regime? If $L$ is fixed and $N$ is allowed to grow then the ratio approaches $L!$. – 2011-05-02
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0Wow, just the slightest change in notation (from lower case to upper) made me misunderstand. That's not a problem with the notation, but a problem with my reading ability. – 2011-05-02
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0Well actually $N$ and $L$ are fixed and less than 100. Thanks for pointing to Stirling's approximation, though. – 2011-05-02