23. From an urn containing $N$ balls, of which $n$ are white and $N-n$ are black, a selection without replacement is made ( $n \geqslant 2$ ). Consider ordered samples of size $M$. What is the probability of events $A_{i, j}$ and $A_{i, j, k}$, where $A_{i, j}$ is the event that white balls are in the $i$-th and $j$-th positions ( $i<j \leqslant M$ ) and $A_{i, j, k}$ is the event that white balls are in the positions with numbers $i, j, k(i<j<k \leqslant M)$?
Solution. The number of elementary outcomes $(N)_{\text {m }}$. Let's calculate the number of elementary outcomes in $A_{i, j}$: on the $i$-th place, we can place $n$ white balls, and on the $j$-th place, the remaining $n-1$ balls. The other $M-2$ places are filled in $(N-2)_{M-2}$ ways. Therefore,

$$
\mathrm{P}\left(A_{i, j}\right)=\frac{n(n-1)(N-2)_{M-2}}{(N)_{M}}=\frac{n(n-1)}{N(N-1)}
$$

For three places, similarly,

$$
\mathbf{P}\left(A_{i, j, k}\right)=\frac{n(n-1)(n-2)}{N(N-1)(N-2)}
$$