How are "(2)" , "(3)" and "(4)" of the "hypergeometric distribution" combined to create the "(5)" or binomial distribution for large N?

Question

Following up to Asymptotic behavior of combinations: approximating Hypergeometric by Binomial.

I'm going to use $k$ instead of $x$, for notational preferences on my part. For $k\leq n$, $$\begin{align} \mathbb{P}\{X=k\}&=\frac{\binom{Np}{k}\binom{Nq}{n-k}}{\binom{N}{n}} \\ &= \frac{(Np)!}{{\color{red}{k!}}(Np-k)!}\cdot\frac{(Nq)!}{{\color{red}{(n-k)!}}(Nq-n+k)!}\cdot\frac{{\color{red}{n!}}(N-n)!}{N!}\\ &= {\color{red}{ \binom{n}{k} }}\cdot \frac{(Np)!}{(Np-k)!}\cdot\frac{(Nq)!}{(Nq-n+k)!}\cdot\frac{(N-n)!}{N!}\tag{1} \\ &={n \choose k} A*B*C\end{align}$$

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).

In "(2)" why are we multiplying by $$\frac{1}{(Np)^k}?$$

In "(3)" why are we multiplying by $$\frac{1}{(Nq)^{n-k}}?$$.

In "(4)" why are we multiplying by $$N^n?$$

Why for "(5)" throw out $A$ and replacing $A$ with the reciprocal of $$\frac{1}{(Np)^{k}}?$$

Why for "(5)" throw out $B$ and replacing that with the reciprocal of $$\frac{1}{(Nq)^{n-k}}?$$

Why for "(5)" throwing out C and replacing C with the reciprocal of $$N^n?$$

Answer 1

Your first three questions are related to what can be done to get limits of $1$ rather than something which depends on $N$.

Your last three questions are related to applying the consequences of the first three.

You cannot say $ \frac{(Np)!}{(Np-k)!} \to (Np)^{k}$ as $N \to \infty$ etc. even if that is what you want to say, so this approach does something more logically coherent

You could use Landau notation to say $ \frac{(Np)!}{(Np-k)!} \sim (Np)^{k}$ as $N \to \infty$ etc. and get to the result that way. The answer you quote spells out the steps in a little more detail.

Answer 2

The following fact is used three times: if $\frac{a}{b} \rightarrow 1$, then $a \rightarrow b$.

In particular, (2) is \begin{align*} \frac{1}{(Np)^{k}}\frac{(Np)!}{(Np-k)!} &\xrightarrow{N\to\infty} 1 \\ \text{so} \qquad (A = {})\ \frac{(Np)!}{(Np-k)!} &\overset{N\to\infty}{\sim} (Np)^{k} \end{align*} and similarly for (3), giving an asymptotic value for $B$, and (4), giving an asymptotic calue for $C$.

In (5), $A$, $B$, and $C$ are replaced with their asymptotic values from (2), (3), and (4).