Example 3 (1987 Singapore Competition Question) 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 have a total of $8 \cdot 7 \cdot 6$ ways to sit in a circle.
On the other hand, if no two people are to sit next to each other, the first person can sit in any position, and the remaining 3 people can sit in every other seat 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 $\frac{24}{8 \cdot 7 \cdot 6}=\frac{1}{14}$.