Verification of poisson approximation to hypergeometric distribution

[kalish1]

How can I verify that

$\lim_{N,M,K \to \infty, \frac{M}{N} \to 0, \frac{KM}{N} \to \lambda} \frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}} = \frac{\lambda^x}{x!}e^{-\lambda}$,

**without** using **Stirling's formula** or the **Poisson approximation to the Binomial**?

I have been stuck on this problem for a while, because I don't know how to divide up the terms and factorials without using the help of prior results!

Any help would be appreciated. Thanks in advance.

 

[Valkarie]

You can verify this limit by using the definition of a binomial coefficient. The binomial coefficient $\binom{n}{k}$ is defined as $\frac{n!}{k!(n-k)!}$. Substituting this definition into your limit, we get: $$\lim_{N,M,K \to \infty, \frac{M}{N} \to 0, \frac{KM}{N} \to \lambda} \frac{\frac{M!(N-M)!}{x!(M-x)!(N-M-x)!}\frac{(N-M)!K!(N-K)!}{(N-M-K)!N!}}{\frac{N!}{K!(N-K)!}} = \frac{\lambda^x}{x!}e^{-\lambda}$$Simplifying the above expression and cancelling out identical terms, we get: $$\lim_{N,M,K \to \infty, \frac{M}{N} \to 0, \frac{KM}{N} \to \lambda} \frac{M!(N-K)!}{x!(M-x)!(N-M-K)!N!} = \frac{\lambda^x}{x!}e^{-\lambda}$$Since $\frac{M}{N} \to 0$ and $\frac{KM}{N} \to \lambda$, we can see that as $N \to \infty$ and $M,K \to \infty$, the terms in the numerator and denominator will cancel out and the limit will become $\frac{\lambda^x}{x!}e^{-\lambda}$ which is the desired result.