Solution. The number of ways in which 3 fixed groups of passengers can be distributed across 10 floors is $A_{10}^{3}$ (the number of arrangements of 10 elements taken 3 at a time). Let's calculate the number of ways to form three groups of 2, 3, and 4 people from 9 people. There are $C_{9}^{2}$ ways to form a group of 2 people, $C_{7}^{3}$ ways to form a group of 3 people, given that the group of 2 people has already been formed. $C_{4}^{4}$ is the number of ways to form a group of 4 people, given that the first two groups have already been formed. Therefore, the total number of ways to form the groups is $C_{9}^{2} \cdot C_{7}^{3} \cdot C_{4}^{4}$. The final result is $A_{10}^{3} \cdot C_{9}^{2} \cdot C_{7}^{3} \cdot C_{4}^{4}=\frac{10!}{4}$.

Answer: $\frac{10!}{4}$.
### 5.088 A lift with 9 passengers can stop at ten floors. Passengers exit in groups of two, three, and four people. In how many ways can this happen?