How to derive hypergeometric distribution pmf via ordered sampling?

Question

So for hypergeometric distribution the set up is as follows. We have $N$ things, $M$ of which are of a type $A$. We pick $n$ things at random without replacement. Let $X$ be the number of things that we pick which are of type $A$. The standard way to derive the probability mass function for X is via unordered sampling without replacement.

I am trying to do this using ordered sampling without replacement instead, but I can't see why the end result should be the same. For my probability $P(X=k)$, I have:

$$P(X=k) = \frac{\frac{M!}{(M-k)!} \times \frac{(N-M)!}{(N-M-n+k)!}}{\frac{N!}{(N-n)!}}$$

and I don't see how this is equal to the standard pmf from unordered sampling approach.

Answer 1

In order to choose an ordered sample of $n$ things which contains exactly $k$ elements of type $A$, you need to

1. Choose which of the $k$ positions in your sample have the type-$A$ objects. There are $\color{red}{\binom nk}$ ways to do this. (This step is missing from the unordered sample version).

2. Fill these $k$ positions with type-$A$ objects. This is done in $\frac{M!}{(M-k)!}$ ways.

3. Fill the remaining $n-k$ positions with type-$B$ objects. This is done in $\frac{(N-M)!}{[(N-M)-(n-k)]!}$ ways.

After dividing by the total number of ordered samples, $\frac{N!}{(N-n)!}$, the probability is $$ P(X=k) = \frac{\color{red}{\binom nk}\times \frac{M!}{(M-k)!} \times \frac{(N-M)!}{(N-M-n+k)!}}{\frac{N!}{(N-n)!}} $$ This now agrees with the formula you get from an unordered analysis.