11. A group of 11 people, including Ivanov and Petrov, are seated around a round table in a random order. Find the probability that there will be 3 people sitting between Ivanov and Petrov.
Solution. Let's denote the event: $A$ - there will be 3 people sitting between Ivanov and Petrov at the table.

We will find the probability of event $A$ by applying formula (1). The numbers $m$ and $n$, which are part of this formula, will be obtained using the formulas of combination theory. There are 11 elements - 11 people in total. In the formation of different combinations (i.e., in the distribution of people around the table), all 11 elements participate, and different combinations differ only in the order of the elements; therefore, these combinations represent permutations of 11 elements. Thus, the total number of outcomes of the experiment $n=P_{11}$. According to formula (2), we find $P_{11}=11$ !.

If the places of Ivanov and Petrov are fixed, for example, seats 1 and 5, and 3 people should sit between them, then the number of different ways to seat people in the remaining 9 seats is $P_{9}=9$ !. The total number of ways favorable to event $A$ is 11 times greater, i.e., it is $m=11 \cdot 9$ !, since there are 11 different ways to seat Ivanov and Petrov so that there are 3 people between them (Ivanov and Petrov can be seated in seats 1 and 5, or 2 and 6, or 3 and 7, etc., up to seats 11 and 4).

The probability of event $A$ is $P(A)=\frac{11 \cdot 9!}{11!}=\frac{1}{10}$, which means it does not depend on the number of people who will sit between Ivanov and Petrov.