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.