2. King Arthur's Round Table. Three knights are randomly seated on chairs arranged around the table at King Arthur's court. What is the probability that there will be an empty chair on both sides of each knight if:

a) there are a total of 7 chairs? (From 6th grade, 1 point).

b) there are a total of $n$ chairs? (From 7th grade, 2 points).
Answer: a) 0.2; b) $\frac{(n-4)(n-5)}{(n-1)(n-2)}$.

Solution. The problem with seven chairs can be solved by enumeration. Let's solve the problem in the general case (part b). For $n<6$, the probability is 0. Consider the case $n \geq 6$. Suppose, for definiteness, that one of these knights is King Arthur himself, and he has taken some seat. Then there are exactly $C_{n-1}^{2}$ ways to seat the other two knights.

Imagine that the king placed his helmet on the next chair to the right, and each knight does the same: places his helmet on the chair to the right of himself. The king and his helmet have already occupied two chairs. There are two pairs of "knight-helmet" left and $n-6$ chairs that are destined to remain empty. In total, there are $n-4$ objects, from which two places for two knights with their helmets need to be chosen. There are $C_{n-4}^{2}$ equally possible ways to do this. The desired probability is

$$
\frac{C_{n-4}^{2}}{C_{n-1}^{2}}=\frac{(n-4)(n-5)}{(n-1)(n-2)}.
$$

In the case $n=7$, we get the answer $\frac{1}{5}$ for part a).