Show that the probability of the sample $S$ is the probability of $S_1 S_2$.

Question

Let $U$ be the population with a size of $N$, and let $n_1$ and $n_2$ be the sizes of two samples $S_1$ and $S_2$, respectively. Let the sample $S = S_1 U S_2$ with size $n = n_1 + n_2$. Show that the probability of selecting this sample is

$\frac{1}{N \choose n_1} \cdot \frac{1}{N-n_1 \choose n_2}= \frac{1}{N\choose n}$.

I've gotten this formula reduced down to $\frac{n_1! n_2! (N-n)!}{N!} = \frac{(n-n_2)!n_2!(N-n)!}{N!}$ but I can't figure out how to get rid of the $n_1!n_2!$ values. I know the final answer needs to be

$\frac{n!(N-n)!}{N!} = \frac{1}{N \choose n}$.

Any tips/suggestions would be greatly appreciated. Thank you in advance.

Answer 1

I get this.

$\frac{1}{N \choose n_1} \cdot \frac{1}{N-n_1 \choose n_2}= \frac{1}{N\choose n} $

is the same as

${N \choose n_1}{N-n_1 \choose n_2}= {N\choose n} $

$\begin{array}\\ {N \choose n_1} {N-n_1 \choose n_2} &=\frac{N!(N-n_1)!}{(N-n_1)!n_1!n_2!(N-n_1-n_2)!}\\ &=\frac{N!}{n_1!n_2!(N-n_1-n_2)!}\\ &=\frac{N!}{n_1!n_2!(N-n)!}\qquad n=n_1+n_2\\ &=\frac{n!N!}{n_1!n_2!n!(N-n)!}\\ &={n \choose n_1}{N \choose n}\\ \end{array} $