Asymptotic behavior of combinations: approximating Hypergeometric by Binomial A form of the hypergeometric distribution is
$$P(X=x)=\frac{\binom{Np}{x}\binom{Nq}{n-x}}{\binom{N}{n}}$$
where 


*

*$N\equiv$ total number of elements of the sample space

*$p\equiv$ probability of choosing a particular type of element

*$q=1-p$

*$x\equiv$ the number of success

*$n\equiv$ the number of trials
It says in the text that as $N\to\infty$, this formula reduces to 
$$P(X=x)=\binom{n}{x}p^xq^{n-x}$$
How is this reduction performed?
I'm asking here rather than in the statistics forum because I'm getting a vibe that it could be more of a mathematical question than a statistical one. (Albeit I'm ready to be shown mistaken)
 A: I'm going to use $k$ instead of $x$, for notational preferences on my part.
For $k\leq n$,
$$\begin{align}
\mathbb{P}\{X=k\}&=\frac{\binom{Np}{k}\binom{Nq}{n-k}}{\binom{N}{n}}
= \frac{(Np)!}{{\color{red}{k!}}(Np-k)!}\cdot\frac{(Nq)!}{{\color{red}{(n-k)!}}(Nq-n+k)!}\cdot\frac{{\color{red}{n!}}(N-n)!}{N!}\\
&= {\color{red}{ \binom{n}{k} }}\cdot \frac{(Np)!}{(Np-k)!}\cdot\frac{(Nq)!}{(Nq-n+k)!}\cdot\frac{(N-n)!}{N!}\tag{1}
\end{align}$$
Now, as $N\to \infty$ and all other parameters are fixed, we have
$$
\frac{1}{(Np)^{k}}\frac{(Np)!}{(Np-k)!} = \frac{(Np)(Np-1)\cdots(Np-k+1)}{(Np)^{k}} \xrightarrow[N\to\infty]{} 1\tag{2}
$$
while 
$$
\frac{1}{(Nq)^{n-k}}\frac{(Nq)!}{(Np-n+k)!} = \frac{(Nq)(Nq-1)\cdots(Nq-n+k+1)}{(Nq)^{n-k}} \xrightarrow[N\to\infty]{} 1\tag{3}
$$
and
$$
N^n\frac{(N-n)!}{N!}= \frac{N^n}{N(N-1)\cdots(N-n+1)} \xrightarrow[N\to\infty]{} 1\tag{4}
$$
so that, combining them all,
$$
\mathbb{P}\{X=k\} \operatorname*{\sim}_{N\to\infty} \binom{n}{k} \frac{(Np)^k(Nq)^{n-k}}{N^n} =\boxed{\binom{n}{k} p^k q^{n-k}} \tag{5}
$$
where $\operatorname*{\sim}_{N\to\infty}$ denotes the standard asymptotic equivalence (Landau notation). 

Last remark. Since the RHS of $(5)$ does not depend on $N$, this turns out to be equivalent (no pun here) to saying that $\lim_{N\to\infty} \mathbb{P}\{X=k\} = \binom{n}{k} p^k q^{n-k}$. 
