$7 \cdot 1$ There are 9 chairs around a round table, and 4 people sit down randomly. What is the probability that no two people are sitting next to each other?
[Solution] The first person can sit in any position, and the remaining three people have a total of $8 \cdot 7 \cdot 6$ ways to sit in a circle.

On the other hand, if it is required that no two people sit next to each other, then the first person can sit in any position, and the remaining 3 people can sit in intervals in a clockwise direction, which has 6 methods. The last extra chair can be in 4 positions, so the total number of methods is $4 \cdot 6=24$.
Therefore, the required probability is $\frac{24}{8 \cdot 7 \cdot 6}=\frac{1}{14}$.