This problem can be solved by considering the constraints: there are 7 people, 3 vehicles, and each vehicle can carry a maximum of 5 people. We need to find the number of ways to distribute the 7 people among the 3 vehicles such that all vehicles are used.

Step-by-step Solution:

1. Distribution of People in Vehicles:

   • We need to distribute 7 people into 3 vehicles such that all vehicles are used. Let's denote the number of people in each vehicle as $x_1$, $x_2$, and $x_3$ where $x_1 + x_2 + x_3 = 7$, and each $x_i \geq 1$ because all vehicles must be used.

   • The problem is to find how many ways we can assign the values to $x_1$, $x_2$, and $x_3$, satisfying $x_1 + x_2 + x_3 = 7$, where each $x_i$ is at least 1, and each $x_i \leq 5$, since each vehicle has a maximum capacity of 5 people.

2. Transforming the Problem:

   • Let’s define new variables $y_1 = x_1 - 1$, $y_2 = x_2 - 1$, and $y_3 = x_3 - 1$, to ensure each vehicle gets at least 1 person.
   • Now, we need to solve $y_1 + y_2 + y_3 = 4$, where $y_1, y_2, y_3 \geq 0$ and $y_1 + 1 \leq 5$, i.e., $y_1 \leq 4$, and similarly for $y_2$ and $y_3$.

3. Counting the Number of Solutions:

   • The equation $y_1 + y_2 + y_3 = 4$ is a classic "stars and bars" problem without any restrictions, and the number of non-negative integer solutions is given by the combination formula: $\binom{4 + 2}{2} = \binom{6}{2} = 15$

4. Distributing the People Among the Vehicles:

   • After determining how many people go into each vehicle, we need to assign specific people to those vehicles. The number of ways to assign 7 people to the vehicles based on the distribution is given by the multinomial coefficient: $\frac{7!}{x_1! x_2! x_3!}$
   • Multiply the two counts together to get the total number of possible distributions.

Thus,