Example 8 There are 5 designated seats, and the 5 people sitting in these 5 seats do not know their designated numbers. When these 5 people randomly sit in the 5 seats, find:
(1) The probability that exactly 3 out of the 5 people sit in their designated seats;
(2) If the probability of these 5 people sitting in their designated seats is not less than $\frac{1}{6}$, what is the maximum number of people who can sit in their designated seats?
Solve: Number 5 people as $1, 2, 3, 4, 5$, and number 5 seats as $1, 2, 3, 4, 5$. From the problem, it is easy to get $n=A_{5}^{5}$.
(1) For 3 people to sit in designated seats, consider it in two steps: The first step is to choose 3 people from 5 to sit in the designated seats, which has $C_{3}^{3}$ ways. The second step is to arrange the remaining two people. For example, if people numbered $1, 2, 3$ sit in their corresponding seats, then people numbered $4, 5$ can only sit in seats numbered $5, 4$, which has 1 way, i.e., $m=C_{5}^{3} \cdot 1$,
Thus, $P=\frac{C_{5}^{3} \cdot 1}{A_{5}^{5}}=\frac{1}{12}$.
(2) From (1), we know that 3 people sitting in designated seats does not meet the requirement. Now consider the case of 2 people: The first step is to choose 2 people from 5 to sit in the designated seats, which has $C_{5}^{2}$ ways. The second step is to arrange the remaining three people. For example, if people numbered $1, 2$ sit in their corresponding seats, then people numbered $3, 4, 5$ can only sit in seats numbered $4, 5, 3$ or $5, 3, 4$, which has 2 ways, i.e., $m=C_{5}^{2} \cdot 2$, Thus, $P=\frac{C_{5}^{2} \cdot 2}{A_{5}^{5}}=\frac{1}{6}$, which meets the requirement, so at most 2 people can sit in their designated seats.