To solve this problem, we need to distribute 7 people across 3 vehicles such that all vehicles are used, with each vehicle accommodating a maximum of 5 people.

Let's break down the problem step by step:

Step 1: Set up the conditions.

• Total number of people = 7
• Number of vehicles = 3
• Each vehicle can hold a maximum of 5 people.
• All vehicles must be used.

Step 2: Partition the 7 people among 3 vehicles.

The number of ways to partition 7 people into 3 non-empty groups (because all vehicles must be used) can be done by first considering possible ways of partitioning and then assigning people.

Possible distributions of people among the vehicles (denoted as [x, y, z] where x, y, and z are the number of people in the vehicles) must satisfy:

• $x + y + z = 7$
• None of $x, y, z$ can exceed 5.

The valid distributions that satisfy these conditions are:

1. [5, 1, 1]
2. [4, 2, 1]
3. [3, 3, 1]

Step 3: Count the number of ways for each distribution.

Now, let's calculate the number of ways to assign 7 people to these distributions.

1. Distribution [5, 1, 1]:

   • Choose 5 people for the first vehicle from 7: $\binom{7}{5} = 21$
   • Choose 1 person for the second vehicle from the remaining 2: $\binom{2}{1} = 2$
   • The last person automatically goes to the third vehicle.

   So, the total number of ways for this distribution is: $21 \times 2 = 42$

2. Distribution [4, 2, 1]:

   • Choose 4 people for the first vehicle from 7: $\binom{7}{4} = 35$
   • Choose 2 people for the second vehicle from the remaining 3: $\binom{3}{2} = 3$
   • The last person automatically goes to the third vehicle.

   So, the total number of ways for this distribution is: $35 \times 3 = 105$

3. Distribution [3, 3, 1]:

   • Choose 3 people for the first vehicle from 7: $\binom{7}{3} = 35$
   • Choose 3 people for the second vehicle from the remaining 4: $\binom{4}{3} = 4$
   • The last person automatically goes to the third vehicle.

   So, the total number of ways for this distribution is: $35 \times 4 = 140$

Step 4: Factor in the arrangement of the vehicles.

Since the vehicles are distinct, the different assignments to the 3 vehicles can be permuted. Therefore, for each partitioning, we multiply by 3! (which is 6) to account for the different ways of assigning groups to the 3 distinct vehicles.

Step 5: Calculate the total number of ways.

Now, we sum the results from each distribution and multiply by 6 (the number of ways to assign the groups to the vehicles): $(42 + 105 + 140) \times 6 = 287 \times 6 = 1722$

Thus, the total number of ways they can visit using all 3 vehicles is 1722.

Conclusion:

None of the options given (3600, 1500, 1650, 1806) match this result exactly, but the closest option is 1806. It might be due to rounding or a slightly different interpretation of the problem, but based on this method, 1722 is the correct calculated value.

Would you like more details or have any further questions?

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Here are five related questions to expand on this topic:

1. How would the answer change if one vehicle could accommodate only 4 people?
2. What if two vehicles were indistinguishable, how would that impact the solution?
3. How would we adjust the solution if the vehicles had different capacities?
4. Can you derive a formula for distributing people into vehicles in general?
5. How does the use of permutation differ from combination in this scenario?

Tip: Always consider the distinctness of objects (like vehicles) when solving combinatorics problems, as it changes whether you use combinations or permutations.