The probability of choosing $(n_1, \space n_2, \space \cdots \space n_k)$ randomly from a set of $(N_1, \space N_2, \space \cdots \space N_k)$ such that for $1\le i\le k, \space i\in \mathbb N$
- $n_i \le N_i$
- $\displaystyle \sum_{1\le i\le k} n_i = n$
- $\displaystyle \sum_{1\le i\le k} N_i = N$
- $p_i = \cfrac {N_i}{n_i}$
is $\cfrac {n!} {n_1!\cdots n_k!}p_1^{n_1}\cdots p_k^{n_k}$ The number of possible outcome is $N^n$ while the number of required outcome is $\cfrac {n!} {n_1!\cdots n_k!} N_1^{n_1}\cdots N_k^{n_k}$.
My question breaks down to proving that $\displaystyle \prod_{1\le i\le k} \left(\! \begin{array}{c} n - (n_1 + n_2 + \cdots +n_{i-1}) \\ n_i \end{array} \!\right) = \cfrac {n!} {n_1!\cdots n_k!}$