To solve this problem, we need to determine the number of ways 7 people can travel to the function in 3 vehicles, each of which can hold up to 5 people, under the condition that all 3 vehicles must be used.

Step 1: Partition the 7 people into 3 groups

The first task is to figure out how to split the 7 people into 3 groups such that no group exceeds 5 people. Since all vehicles must be used, no group can have zero members. Therefore, we must partition 7 people into 3 non-empty groups where each group has a size between 1 and 5.

The only feasible partition that satisfies these constraints (since each vehicle can hold at most 5 people) is:

• One group of 3 people
• Two groups of 2 people each

Step 2: Calculate the number of ways to assign people to these groups

Now, we must calculate the number of ways to assign the 7 people to these groups. This can be broken into two parts:

1. Choose the 3 people who will be in the group of 3:
   This can be done in $\binom{7}{3}$ ways, where $\binom{7}{3}$ is the number of ways to choose 3 people from 7. The formula for combinations is:

   $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

2. Choose the 2 people for the first group of 2:
   After selecting 3 people, 4 people remain. We choose 2 of them to be in one of the groups of 2. This can be done in $\binom{4}{2}$ ways:

   $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

3. The remaining 2 people automatically go to the last group of 2:
   There is only 1 way to assign the last 2 people to the last group.

Thus, the total number of ways to partition the 7 people into one group of 3 and two groups of 2 is:

$\binom{7}{3} \times \binom{4}{2} \times 1 = 35 \times 6 \times 1 = 210$

Step 3: Assign groups to vehicles

Since we have 3 groups and 3 vehicles, we need to assign each group to one of the 3 vehicles. There are $3!$ ways to do this:

$3! = 3 \times 2 \times 1 = 6$

Step 4: Total number of ways

The total number of ways to assign 7 people into 3 groups, and then assign each group to a vehicle, is:

$210 \times 6 = 1260$

Thus, the total number of ways the 7 people can visit the function using all 3 vehicles is 1,260.

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Would you like more details or have any questions? Here are five related questions to explore further:

1. How does the problem change if only 2 vehicles are required?
2. What if the maximum capacity of each vehicle was reduced to 4 persons?
3. How would the solution change if there were 8 people instead of 7?
4. What happens if we allow some vehicles to remain unused?
5. Can you calculate the number of ways if one vehicle can only hold exactly 1 person?

Tip: The factorial function $n!$ is a crucial concept in combinatorics, representing the product of all integers from 1 to $n$. It’s used often in problems involving counting arrangements or combinations.