We are given that 7 people are attending a marriage function using 3 different vehicles, each with a maximum capacity of 5 people. We need to determine how many ways these 7 people can travel in such a way that all 3 vehicles are used.

Step-by-step solution:

1. Total people and vehicles:

   • We have 7 people.
   • There are 3 vehicles, and each vehicle can carry up to 5 people.

2. Condition: All 3 vehicles must be used. This means none of the vehicles can be empty. The 7 people must be distributed across the 3 vehicles in such a way that no vehicle is left out.

3. Valid partition of people: We need to find a way to partition 7 people into 3 non-empty groups where the number of people in each group represents how many are in each vehicle.

   Since each vehicle must have at least 1 person and all vehicles need to be used, the possible partitions of 7 into 3 groups are:

   • (5, 1, 1) → One vehicle has 5 people, and the other two vehicles have 1 person each.
   • (4, 2, 1) → One vehicle has 4 people, another has 2, and the last one has 1 person.
   • (3, 3, 1) → Two vehicles have 3 people, and the last one has 1 person.

   These are the only valid partitions.

4. Calculating the number of ways for each partition:

   For each partition, we calculate the number of ways to assign people to vehicles:

   • (5, 1, 1):

     • Choose 5 people out of 7 for the first vehicle: $\binom{7}{5} = \binom{7}{2} = 21$.
     • Out of the remaining 2 people, choose 1 for the second vehicle: $\binom{2}{1} = 2$.
     • The last person automatically goes into the third vehicle.
     • Now, since the vehicles are distinct, the number of ways to assign the 3 groups to the 3 vehicles is $\frac{3!}{2!} = 3$ (since two of the vehicles have 1 person each, they are indistinguishable).
     • Total for this case: $21 \times 2 \times 3 = 126$.

   • (4, 2, 1):

     • Choose 4 people out of 7 for the first vehicle: $\binom{7}{4} = \binom{7}{3} = 35$.
     • Out of the remaining 3 people, choose 2 for the second vehicle: $\binom{3}{2} = 3$.
     • The last person automatically goes into the third vehicle.
     • All vehicles are distinct, so no further division is needed. Number of ways to assign groups to vehicles is $3!$.
     • Total for this case: $35 \times 3 \times 6 = 630$.

   • (3, 3, 1):

     • Choose 3 people out of 7 for the first vehicle: $\binom{7}{3} = 35$.
     • Out of the remaining 4 people, choose 3 for the second vehicle: $\binom{4}{3} = 4$.
     • The last person automatically goes into the third vehicle.
     • Now, since two of the vehicles have the same number of people, they are indistinguishable. The number of ways to assign the groups to vehicles is $\frac{3!}{2!} = 3$.
     • Total for this case: $35 \times 4 \times 3 = 420$.

5. Total number of ways: Now, sum up the results from all three cases: $126 + 630 + 420 = 1176$

Thus, the total number of ways the 7 people can travel using all 3 vehicles is 1176.

---

Would you like more details or have any questions?

Here are 5 questions to explore further:

1. What if the vehicles had different capacities? How would the solution change?
2. How do you calculate combinations like $\binom{n}{k}$?
3. What if the vehicles were indistinguishable, how would the number of ways change?
4. What if we had more than 3 vehicles but still wanted to use exactly 3?
5. How do you approach such problems in combinatorics in general?

Tip: For problems involving distribution of people into groups with conditions, partitioning and combinations play a key role.