5
$\begingroup$

Any ideas on finding a good estimate/approximation for $\frac AB$ where $A = N^L$ and $B = {N+L\choose N}$?

  • 2
    You could try applying Stirling on the factorials implicit in the binomial coefficient...2011-05-02
  • 1
    I don't understand the notations $N^L$ and $C_{N+L}^N$. What do those mean?2011-05-02
  • 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
  • 1
    In what regime? If $L$ is fixed and $N$ is allowed to grow then the ratio approaches $L!$.2011-05-02
  • 0
    Wow, 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
  • 0
    Well actually $N$ and $L$ are fixed and less than 100. Thanks for pointing to Stirling's approximation, though.2011-05-02

3 Answers 3