Sampling distribution of random sample I am currently working on the following question: Suppose you have a finite population $P$ of size $N$. We select a sample $S_1$ using a random sampling without replacement of size $n_1$. Then we select a sample $S_2$ from $P-S_1$ using random sampling without replacement of size $n_2$. 
Then we define $S= S_1 \cup S_2$. What is the sampling distribution of $S$?
$My \ answer$: 
There are $\binom{N}{n_1}$ possible ways to have a random sample $S_1$. Then there are $\binom{N-n_1}{n_2}$ possible ways to have a random sample $S_2$. Hence the number of possibilities for a random sample $S$ equals $\binom{N}{n_1}\binom{N-n_1}{n_2}$. Now we have counted possible outcomes multiple times. There are $\binom{n_1+n_2}{n_1}$ possible ways to have a specific selection. Hence the total numer of outcomes equals $\binom{N}{n_1}\binom{N-n_1}{n_2}/\binom{n_1+n_2}{n_1} = \binom{N}{n_1+n_2}$.
Therefore the probability of a specific sample equals $[\binom{N}{n_1+n_2}]^{-1}$
Is my reasoning correct? Thanks in advance!

Yes, but more directly you could say that this probability equals:$$\binom{N}{n_1+n_2}^{-1}$$
There is no essential difference between taking two consecutive samples of sizes $n_1$, $n_2$ respectively that are afterwards joined and one sample of size $n_1+n_2$.
