Hypergeometric distribution. I read that we can use hypergeometric distribution for finding the probability for without replacement cases because the probability of a particular event changes on every trial and binomial distribution fails.
I know how to solve questions using hypergeometric distribution but I don't understand how does using combinatorics for finding the probability helps in without replacement cases. 
Can someone please tell me why hypergeometric distribution works for cases without replacement?
 A: Let´s say you have $R$ (red) and $B$ (blue) elements, where $B+R=N$. Now you want to draw $r \in R $ elements and $b\in B$ without replacement. Again $r+b=n$
The number of ways to order $r$ drawn elements out of $R$ is $\binom{R}{r}$. For instance, if $R=4$ and $r=2$, then you have the following combinations to draw the two red elements out of $4$ red elements
$(1,2);(1;3);(1;4);(2,3);(2,4);(3,4)$ 
Thus there are $6$ combinations which is equal to $\binom{R}{r}=\binom{4}{2}=6$
The same calculation can be done for the blue elements. To get the  numbers to draw $r$ and $b$ elements (favorable cases, $\color{green}{F}$) the binomial coefficients has to be multiplied. We use that $B=N-R$ and $b=n-r$.
$$F=\binom{R}{r}\cdot \binom{N-R}{n-r}$$
Now we divide the term above by the number of all possible cases, $\color{violet}{P}$. Here you draw $r+b=n$ elements out of $N=R+B$ cases. You don´t care how much of the $n$ elements are blue or red, only $r\leq R$ and $b\leq B$ Thus the probability to draw $r$ and $b$ elements out of R and B elements is 
$$\frac{\color{green}{F}}{\color{violet}{P}}=\frac{\binom{R}{r}\cdot \binom{N-R}{n-r}}{\binom{N}{n}}$$
