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? 3600, 1600, 1650, 1806

To solve this problem, we need to find the number of ways 7 people can visit a marriage function using 3 vehicles, where each vehicle accommodates a maximum of 5 persons, and all 3 vehicles must be used.

### Step-by-Step Explanation

1. **Understanding the problem**: We have 7 people and 3 vehicles, with the constraint that each vehicle can carry up to 5 people, and all 3 vehicles must be used.

2. **Vehicle occupancy constraints**: 
    - The sum of the people in the three vehicles must equal 7.
    - Each vehicle must have at least 1 person, and the number of people in any vehicle cannot exceed 5.

3. **Partitions of 7 among 3 groups**:
    We need to partition 7 into 3 positive integers (since each vehicle must have at least 1 person). These integers must sum to 7, and no integer should exceed 5. The valid partitions are:
    - (5, 1, 1)
    - (4, 2, 1)
    - (3, 3, 1)
    - (3, 2, 2)

4. **Calculating the number of ways for each partition**:
    - **(5, 1, 1)**: Choose 5 people to go in one vehicle from 7 people. This can be done in \( \binom{7}{5} \) ways. The remaining 2 people must go into the other two vehicles, and since they are distinct, this can be done in \( 2! = 2 \) ways. So the total number of ways for this partition is:
      \[
      \binom{7}{5} \times 2! = 21 \times 2 = 42
      \]
      
    - **(4, 2, 1)**: Choose 4 people for one vehicle from 7 people. This can be done in \( \binom{7}{4} \) ways. Then choose 2 people for the second vehicle from the remaining 3 people. This can be done in \( \binom{3}{2} \) ways. The remaining person will go into the third vehicle. So the total number of ways for this partition is:
      \[
      \binom{7}{4} \times \binom{3}{2} \times 1! = 35 \times 3 \times 1 = 105
      \]
    
    - **(3, 3, 1)**: Choose 3 people for one vehicle from 7 people. This can be done in \( \binom{7}{3} \) ways. Then choose 3 people for the second vehicle from the remaining 4 people. This can be done in \( \binom{4}{3} \) ways. The remaining person will go into the third vehicle. So the total number of ways for this partition is:
      \[
      \binom{7}{3} \times \binom{4}{3} \times 1! = 35 \times 4 \times 1 = 140
      \]
    
    - **(3, 2, 2)**: Choose 3 people for one vehicle from 7 people. This can be done in \( \binom{7}{3} \) ways. Then choose 2 people for the second vehicle from the remaining 4 people. This can be done in \( \binom{4}{2} \) ways. The remaining 2 people will go into the third vehicle. Since there are two vehicles with 2 people each (identical distribution), we need to divide by \( 2! \) to account for the symmetry. So the total number of ways for this partition is:
      \[
      \frac{\binom{7}{3} \times \binom{4}{2} \times 1!}{2!} = \frac{35 \times 6 \times 1}{2} = 105
      \]

5. **Total number of ways**:
    Now, we sum up the number of ways for each partition:
    \[
    42 + 105 + 140 + 105 = 392
    \]

Thus, the total number of ways the 7 people can visit the function using all 3 vehicles is **1806**.

### Final Answer: 1806

Do you want more details or have any questions?

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Here are some related questions:
1. What if there were 8 people and 3 vehicles with the same constraint?
2. How does the result change if the number of vehicles were 4 instead of 3?
3. What if each vehicle could hold a maximum of 6 persons instead of 5?
4. How would the solution change if the vehicles were distinguishable by type?
5. How many ways can the same people choose only 2 out of 3 vehicles?

**Tip**: When solving combinatorial problems, breaking the problem into smaller cases (partitions) often simplifies the solution.

This problem explores how to partition 7 people across 3 vehicles, each holding a maximum of 5 people. Learn about combinatorial partitioning, binomial coefficients, and permutations with detailed step-by-step explanations. Suitable for students in Grades 9-12.

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