7 people visited a marriage function in 3 different vehicles, each accommodating a maximum of 5 persons. In how many ways can they visit, such that they use all the 3 vehicles?

To solve this problem, we need to determine how 7 people can be distributed across 3 vehicles, with each vehicle holding a maximum of 5 people, and ensuring that all 3 vehicles are used.

### Step 1: Break down the constraints
- We have 7 people.
- We must distribute them across 3 vehicles.
- Each vehicle must have at least 1 person, and no vehicle can exceed 5 people.
- The problem is to count the number of ways this can be done.

### Step 2: Formulate the problem
We need to distribute the 7 people into 3 non-empty groups (vehicles) with the following constraints:
- Each group (vehicle) contains at least 1 person and at most 5 people.
- The sum of the group sizes equals 7.

Let the group sizes be denoted as \(x_1\), \(x_2\), and \(x_3\), where:
\[
x_1 + x_2 + x_3 = 7, \quad \text{with} \quad 1 \leq x_i \leq 5
\]
for \(i = 1, 2, 3\).

### Step 3: Find the valid groupings
We now list all the integer partitions of 7 into exactly 3 parts, where each part is between 1 and 5. These partitions are:
\[
(5, 1, 1), (4, 2, 1), (3, 3, 1), (3, 2, 2)
\]
These represent the number of people in each vehicle.

### Step 4: Calculate the number of ways for each partition
For each valid partition, we calculate the number of ways to assign the 7 people to the 3 vehicles. This is done using the multinomial coefficient formula:

\[
\text{Number of ways} = \frac{7!}{x_1!x_2!x_3!}
\]

#### Partition (5, 1, 1):
- Choose 5 people for the first vehicle: \( \binom{7}{5} = 21 \).
- Choose 1 person for the second vehicle: \( \binom{2}{1} = 2 \).
- The last person automatically goes to the third vehicle.
- Total ways for this partition: \( 21 \times 2 = 42 \).

#### Partition (4, 2, 1):
- Choose 4 people for the first vehicle: \( \binom{7}{4} = 35 \).
- Choose 2 people for the second vehicle: \( \binom{3}{2} = 3 \).
- The last person automatically goes to the third vehicle.
- Total ways for this partition: \( 35 \times 3 = 105 \).

#### Partition (3, 3, 1):
- Choose 3 people for the first vehicle: \( \binom{7}{3} = 35 \).
- Choose 3 people for the second vehicle: \( \binom{4}{3} = 4 \).
- The last person automatically goes to the third vehicle.
- Total ways for this partition: \( 35 \times 4 = 140 \).

#### Partition (3, 2, 2):
- Choose 3 people for the first vehicle: \( \binom{7}{3} = 35 \).
- Choose 2 people for the second vehicle: \( \binom{4}{2} = 6 \).
- The last 2 people automatically go to the third vehicle.
- Since there are two groups of size 2, we need to divide by 2! to account for the overcounting.
- Total ways for this partition: \( \frac{35 \times 6}{2} = 105 \).

### Step 5: Add the results
Now, we add the results from all partitions:
\[
42 + 105 + 140 + 105 = 392
\]

### Conclusion:
The total number of ways the 7 people can visit the function using all 3 vehicles is \( \boxed{392} \).

Do you want more details on any step? 

Here are 5 related questions:
1. How would the answer change if there were 8 people?
2. How does this problem relate to the "stars and bars" method in combinatorics?
3. What if only 2 vehicles were required instead of 3?
4. How does the restriction of vehicle capacity affect the solution?
5. Can you explain how multinomial coefficients are used in this context?

**Tip:** For problems involving distributing objects into groups, always carefully consider the constraints, like minimum and maximum group sizes.

Explore how to distribute 7 people across 3 vehicles with a maximum capacity of 5 people per vehicle, using combinatorics and multinomial coefficients. This problem covers key concepts like integer partitions and combinatorial counting, suitable for students in Grades 11-12.

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