8. There are 9 representatives, each from different countries, with 3 people from each country. They randomly sit at a round table with 9 chairs. The probability that each representative is seated next to at least one representative from another country is $\qquad$ .
8. $\frac{41}{56}$.
9 The number of ways to arrange the chairs by country is the multinomial coefficient $\frac{9!}{3!3!3!}=1680$. The required probability is 1 minus the probability that at least three representatives from one country sit together.

Let the sets of seating arrangements where the representatives from the first, second, and third countries sit together be $A, B, C$ respectively. By the principle of inclusion-exclusion and the symmetry of sets $A, B, C$, we have
$$
\begin{aligned}
|A \cup B \cup C| & =(|A|+|B|+|C|)-(|A \cap B|+|A \cap C|+|B \cap C|)+|A \cap B \cap C| \\
& =3|A|-3|A \cap B|+|A \cap B \cap C| .
\end{aligned}
$$

To determine the number of elements in $A$, we can choose a chair for the rightmost representative from the first country. The two chairs to the left of this chair are also occupied by representatives from the first country. The remaining 6 chairs are then arranged for the representatives from the other two countries. Thus, $|A|=9 \mathrm{C}_{6}^{3}$.

To determine the number of elements in $A \cap B$, we can choose a chair for the rightmost representative from the first country. The two chairs to the left of this chair are also occupied by representatives from the first country. From the remaining 6 chairs, the rightmost representative from the second country can choose from 4 chairs, and all positions of the representatives are then determined. Thus, $|A \cap B|=9 \times 4$.

To determine the number of elements in $A \cap B \cap C$, we can choose a chair for the rightmost representative from the first country. The two chairs to the left of this chair are also occupied by representatives from the first country. From the remaining 6 chairs, the seating arrangements for the representatives from the other two countries have 2 possibilities. Thus, $|A \cap B \cap C|=9 \times 2$.
In summary, $|A \cup B \cup C|=3\left(9 \mathrm{C}_{6}^{3}\right)-3(9 \times 4)+9 \times 2=450$.
The required probability is $1-\frac{|A \cup B \cup C|}{1680}=1-\frac{450}{1680}=\frac{41}{56}$.