8. (SPA 2) In a test, $3 n$ students participate, who are located in three rows of $n$ students in each. The students leave the test room one by one. If $N_{1}(t), N_{2}(t), N_{3}(t)$ denote the numbers of students in the first, second, and third row respectively at time $t$, find the probability that for each $t$ during the test,  
$$ \left|N_{i}(t)-N_{j}(t)\right|<2, \quad i \neq j, \quad i, j=1,2, \ldots $$
8. Situations in which the condition of the statement is fulfilled are the following: $S_{1}: N_{1}(t)=N_{2}(t)=N_{3}(t)$ $S_{2}: N_{i}(t)=N_{j}(t)=h, N_{k}(t)=h+1$, where $(i, j, k)$ is a permutation of the set $\{1,2,3\}$. In this case the first student to leave must be from row $k$. This leads to the situation $S_{1}$. $S_{3}: N_{i}(t)=h, N_{j}(t)=N_{k}(t)=h+1,((i, j, k)$ is a permutation of the set $\{1,2,3\})$. In this situation the first student leaving the room belongs to row $j$ (or $k$) and the second to row $k$ (or $j$). After this we arrive at the situation $S_{1}$. Hence, the initial situation is $S_{1}$ and after each triple of students leaving the room the situation $S_{1}$ must recur. We shall compute the probability $P_{h}$ that from a situation $S_{1}$ with $3 h$ students in the room $(h \leq n)$ one arrives at a situation $S_{1}$ with $3(h-1)$ students in the room: 
$$ P_{h}=\frac{(3 h) \cdot(2 h) \cdot h}{(3 h) \cdot(3 h-1) \cdot(3 h-2)}=\frac{3!h^{3}}{3 h(3 h-1)(3 h-2)} $$
Since the room becomes empty after the repetition of $n$ such processes, which are independent, we obtain for the probability sought 
$$ P=\prod_{h=1}^{n} P_{h}=\frac{(3!)^{n}(n!)^{3}}{(3 n)!} $$