7 people visited a marriage function in 3 different vehicles, each accomodating 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 many ways 7 people can visit the marriage function using 3 vehicles, such that all vehicles are used and each vehicle accommodates a maximum of 5 persons.

### Step-by-Step Solution:

1. **Vehicle Capacity Constraints**:
   - Each vehicle can carry at most 5 people.
   - There are 7 people total, and all three vehicles must be used.
   
2. **Possible Distributions of People Across Vehicles**:
   To use all three vehicles, we need to distribute the 7 people such that:
   - The number of people in each vehicle must be within the range of 1 to 5 people (due to the capacity limit).
   
   Considering these conditions, the possible distributions of people among the 3 vehicles are:
   - Vehicle 1: 5 people, Vehicle 2: 1 person, Vehicle 3: 1 person (5, 1, 1)
   - Vehicle 1: 4 people, Vehicle 2: 2 people, Vehicle 3: 1 person (4, 2, 1)
   - Vehicle 1: 3 people, Vehicle 2: 3 people, Vehicle 3: 1 person (3, 3, 1)

3. **Calculating the Number of Ways for Each Distribution**:
   For each distribution, we need to compute how many ways we can assign people to vehicles.

   - **Case 1: Distribution (5, 1, 1)**:
     - Choose 5 people to go in the first vehicle: \( \binom{7}{5} = 21 \)
     - Choose 1 person to go in the second vehicle: \( \binom{2}{1} = 2 \)
     - The remaining 1 person will go in the third vehicle.
     - Total number of ways for this distribution: \( 21 \times 2 = 42 \)

   - **Case 2: Distribution (4, 2, 1)**:
     - Choose 4 people to go in the first vehicle: \( \binom{7}{4} = 35 \)
     - Choose 2 people to go in the second vehicle: \( \binom{3}{2} = 3 \)
     - The remaining 1 person will go in the third vehicle.
     - Total number of ways for this distribution: \( 35 \times 3 = 105 \)

   - **Case 3: Distribution (3, 3, 1)**:
     - Choose 3 people to go in the first vehicle: \( \binom{7}{3} = 35 \)
     - Choose 3 people to go in the second vehicle: \( \binom{4}{3} = 4 \)
     - The remaining 1 person will go in the third vehicle.
     - Total number of ways for this distribution: \( 35 \times 4 = 140 \)

4. **Total Number of Ways**:
   Now, we sum up the total number of ways from all possible distributions:
   \[
   42 \, (\text{for } (5,1,1)) + 105 \, (\text{for } (4,2,1)) + 140 \, (\text{for } (3,3,1)) = 287
   \]

Thus, the total number of ways in which 7 people can visit using all three vehicles is **287**.

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Would you like any further clarifications? Here are some related questions you might find interesting:

1. How many ways can 8 people visit using 3 vehicles with the same conditions?
2. What if one of the vehicles had a capacity of 6 instead of 5? How would this change the result?
3. How many ways can they visit if there were no restrictions on the number of vehicles they used?
4. How would the answer change if one vehicle must have at least 3 people?
5. What is the total number of ways to seat 7 people in 3 vehicles if the vehicles can accommodate all of them?

**Tip**: When dealing with combinatorial problems, consider using the binomial coefficient formula \( \binom{n}{k} \), which represents the number of ways to choose \( k \) objects from \( n \) without regard to the order.

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?

Learn how to solve a combinatorics problem where 7 people are distributed across 3 vehicles, each with a capacity of 5. This detailed step-by-step solution covers binomial coefficients, permutations, and combinations, providing insight into how to distribute groups under constraints. Suitable for students in Grades 9-12.

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