How to calculate $P(X=x,Y =y,Z=z)$ for $x, y, z$ non-negative integers with $x+y+z=m$ I'm currently working on a question from BH. 
A certain country has three political parties, denoted by $A, B$, and $C$. Each adult in the country is a member of exactly one of the three parties. There are $n$ adults in the country, consisting of $n_A$ members of party $A$, $n_B$ members of party $B$, and $n_C$ members of party $C$, where $n_A, n_B , n_C$ are positive integers with $n_A + n_B + n_C = n$.
A simple random sample of size m is chosen from the adults in the country (the sampling is done without replacement, and all possible samples of size m are equally likely). Let $X$ be the number of members of party $A$ in the sample, $Y$ be the number of members of party $B$ in the sample, and $Z$ be the number of members of party $C$ in the sample.
(a) Find $P(X=x,Y =y,Z=z)$, for $x,y,z$ non negative integers with $x+y+z=m$.
(b) Find $E(X)$.
(c) Find $\mathrm{Var}(X)$, and briefly explain why your answer makes sense in the extreme cases $m = 1$ and $m = n$.
Currently for part a) I have that there are in total $nCm$ choices with $$\frac{nCn_a \cdot n_aCx + (n-x)Cn_b \cdot n_bCy + (n-x-y)Cn_c \cdot n_cCy}{nCm}$$
Pardon my formatting. I think this question has something to do with a HyperGeometric distribution, but I'm having trouble connect that with Binomial and then to Poisson. 
 A: HINT
First of all, you can format "n choose m" several ways, but the way I prefer is "{n \choose m}".  If you put all that inside mathjax (i.e. between \$ \$ signs), you get ${n \choose m}$.
(a) I think your answer isn't right.  There are of course ${n \choose m}$ subsets, so the denominator is correct.  For the numerator, how many ways are there to choose $x$ out of $n_A$, and $y$ out of $n_B$, and $z$ out of $n_C$?  It is simply ${n_A \choose x} {n_B \choose y}{n_C \choose z}$ and that should be the numerator.
Your proposed answer starts with ${n \choose n_A}{n_A \choose x}$.  This counts the number of ways to choose the party $A$ voters, and then to choose $x$ among them.  However, as far as the random sampling is concerned, it doesn't matter who else belongs to party $A$ as long as $x$ are chosen.  So you are over-counting.
E.g. imagine the case of $n_A=n_B=n_C=2$ and $x=y=z=1$.  There are clearly $2^3 =8$ ways to pick the desired subset.  Or even simpler: $x=2, y=z=0$ and there's only $1$ such subset.  In both cases your numerator would over-count by a lot.
(b) The fastest way is to use linearity of expectations.
(c) I'm not sure there's an easy way to do this one...  but it may help that you're really only dealing with a binomial distribution (either a chosen voter belongs to $A$, or not).
Hope this helps...?
