I'm having trouble understand the part of the pmf for the Hypergeometric Distribution highlighted in green:
$$\Pr[X = k] = \frac{\dbinom{m}{k}\!\!\color{green}{\dbinom{N-m}{n-k}}}{\dbinom{N}{n}}$$
If you have chosen k "Type 1" objects from a maximum possible m, then the number of ways of choosing this is ${m\choose k}$. But why do you then require the part highlighted in green? I would have thought this would have already been accounted for given that it is implied by ${m\choose k}$.
In other words why is $P(X=k)\neq \displaystyle \frac{{m\choose k}}{{N\choose n}}$
Say you have a class of $N = 12$ students, of which $m = 5$ are boys. That means $N - m = 7$ are girls. How many ways can you choose $n = 6$ students such that $k = 2$ are boys? Clearly, there are $\binom{5}{2}$ ways to choose the two boys, but then there are also $\binom{7}{4}$ ways to choose the $6-2 = 4$ girls. The fact that you also have to choose $4$ girls out of the $7$ is implicit in the requirement that you choose $6$ total students. Since the choice of which boys to select is independent of the choice of which girls to select, the total number of desired outcomes is the product of these two binomial coefficients, $\binom{5}{2}\binom{7}{4}$. Now, the probability of selecting $k = 2$ boys out of $n = 6$ randomly chosen students in the class (without replacement) is the desired outcomes divided by the possible outcomes, which is $\binom{12}{6}$. Hence the desired probability is $$\frac{\binom{5}{2}\binom{7}{4}}{\binom{12}{6}}.$$