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?

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**.

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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.

Explore how to distribute 7 people across 3 distinct vehicles, with each vehicle accommodating up to 5 people. This detailed solution covers partitioning and combinatorics concepts, including key formulas for combinations and permutations. Ideal for students in Grades 10-12.

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