To solve this problem, we need to determine how many ways 7 people can travel in 3 vehicles with the condition that all vehicles are used, and each vehicle can accommodate a maximum of 5 people.

Step 1: Partitioning the 7 people into 3 non-empty groups

We first need to partition the 7 people into 3 groups, representing the number of people in each vehicle. These groups cannot be empty, and the maximum size for each group is 5.

Let the number of people in the vehicles be denoted by $(x_1, x_2, x_3)$, where $x_1 + x_2 + x_3 = 7$ and each $x_i \geq 1$, meaning every vehicle must have at least 1 person. The possible valid groupings (which satisfy $x_i \leq 5$) are:

• $(5, 1, 1)$
• $(4, 2, 1)$
• $(3, 3, 1)$
• $(3, 2, 2)$

Step 2: Counting the arrangements for each partition

For each partition, we will calculate the number of ways to assign people to these groups.

1. For the partition $(5, 1, 1)$:

   • Choose 5 people to be in the first vehicle from 7: $\binom{7}{5} = 21$
   • Choose 1 person for the second vehicle from the remaining 2 people: $\binom{2}{1} = 2$
   • The last person goes into the third vehicle.
   • There are $3!$ (since the vehicles are distinguishable) different ways to assign the three groups to the three vehicles.

   So, the total number of ways for this partition is: $21 \times 2 \times 1 \times 3! = 21 \times 2 \times 6 = 252$

2. For the partition $(4, 2, 1)$:

   • Choose 4 people for the first vehicle from 7: $\binom{7}{4} = 35$
   • Choose 2 people for the second vehicle from the remaining 3: $\binom{3}{2} = 3$
   • The last person goes into the third vehicle.
   • Again, multiply by $3!$ to account for the distinguishable vehicles.

   The total number of ways for this partition is: $35 \times 3 \times 1 \times 3! = 35 \times 3 \times 6 = 630$

3. For the partition $(3, 3, 1)$:

   • Choose 3 people for the first vehicle from 7: $\binom{7}{3} = 35$
   • Choose 3 people for the second vehicle from the remaining 4: $\binom{4}{3} = 4$
   • The last person goes into the third vehicle.
   • Multiply by $3!$ to account for distinguishable vehicles.

   The total number of ways for this partition is: $35 \times 4 \times 1 \times 3! = 35 \times 4 \times 6 = 840$

4. For the partition $(3, 2, 2)$:

   • Choose 3 people for the first vehicle from 7: $\binom{7}{3} = 35$
   • Choose 2 people for the second vehicle from the remaining 4: $\binom{4}{2} = 6$
   • The last 2 people go into the third vehicle.
   • Multiply by $3!$ to account for distinguishable vehicles.

   The total number of ways for this partition is: $35 \times 6 \times 1 \times 3! = 35 \times 6 \times 6 = 1260$

Step 3: Adding all possibilities

Now, sum up the possibilities for each partition: $252 + 630 + 840 + 1260 = 2982$

Thus, the total number of ways the 7 people can visit the marriage function using all 3 vehicles is 2982.

Would you like further clarification or additional details?

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Here are 5 related questions to deepen your understanding:

1. How would the answer change if each vehicle could accommodate only 4 people?
2. What if there were 4 vehicles, but only 3 must be used?
3. How many ways would there be if the vehicles were indistinguishable?
4. Can this problem be generalized for any number of people and vehicles?
5. How would the answer change if each vehicle had a different seating capacity?

Tip: When partitioning people into groups, start by calculating all possible partitions that satisfy given conditions and then apply combinatorics principles to count arrangements.