We are given that 7 people need to be accommodated in 3 different vehicles, each with a maximum capacity of 5 people. The question asks how many ways this can be done while ensuring all 3 vehicles are used.

Step 1: Partitioning into groups

Since all three vehicles must be used, we need to divide 7 people into 3 non-empty groups such that each group represents the number of people in a vehicle. Let the number of people in the three groups be $x_1$, $x_2$, and $x_3$, where:

$x_1 + x_2 + x_3 = 7$

And each group must satisfy:

$1 \leq x_1, x_2, x_3 \leq 5$

Thus, the possible partitions of 7 into 3 positive integers, with each integer less than or equal to 5, are:

• $(5, 1, 1)$
• $(4, 2, 1)$
• $(3, 3, 1)$
• $(3, 2, 2)$

Step 2: Calculating the number of ways for each partition

For each partition, we will calculate the number of ways to assign people to the vehicles.

Partition $(5, 1, 1)$:

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

Thus, the total number of ways for this partition is:

$21 \times 2 = 42 \text{ ways}$

Partition $(4, 2, 1)$:

• Choose 4 people to go into the first vehicle: $\binom{7}{4} = 35$ ways.
• Choose 2 people for the second vehicle: $\binom{3}{2} = 3$ ways.
• The remaining 1 person goes into the third vehicle automatically.

Thus, the total number of ways for this partition is:

$35 \times 3 = 105 \text{ ways}$

Partition $(3, 3, 1)$:

• Choose 3 people to go into the first vehicle: $\binom{7}{3} = 35$ ways.
• Choose 3 people for the second vehicle from the remaining 4 people: $\binom{4}{3} = 4$ ways.
• The remaining 1 person goes into the third vehicle automatically.

Thus, the total number of ways for this partition is:

$35 \times 4 = 140 \text{ ways}$

Partition $(3, 2, 2)$:

• Choose 3 people to go into the first vehicle: $\binom{7}{3} = 35$ ways.
• Choose 2 people for the second vehicle from the remaining 4 people: $\binom{4}{2} = 6$ ways.
• The remaining 2 people go into the third vehicle automatically.

Thus, the total number of ways for this partition is:

$35 \times 6 = 210 \text{ ways}$

Step 3: Summing all the possibilities

Now, we add up the total number of ways for each partition:

$42 + 105 + 140 + 210 = 497 \text{ ways}$

Thus, the total number of ways the 7 people can visit using all 3 vehicles is $\boxed{497}$.

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Would you like further details or clarifications?

Here are 5 related questions:

1. How do binomial coefficients work in calculating combinations?
2. What if the vehicles had different capacities?
3. Can we generalize this problem to any number of people and vehicles?
4. What happens if we relax the condition of using all vehicles?
5. How would the solution change if one vehicle had a capacity less than 5?

Tip: Partitioning problems often rely on careful counting and using combinations to select groups. Make sure to explore all valid group configurations!