seven people visited a marriage function in three different vehicles, each accommodating, a maximum of 5% and how many ways can they visit such that they use all three vehicles

This problem can be solved using combinatorics, specifically the **stars and bars** method combined with the inclusion-exclusion principle to ensure all three vehicles are used.

### Problem Restatement:
- **7 people** need to visit a marriage function.
- There are **3 vehicles**, each with a **maximum capacity of 5 people**.
- We need to find how many ways these 7 people can visit using **all three vehicles**.

### Key Constraints:
- **Each vehicle must be used**, and each can carry at most 5 people.
  
### Step 1: Understand the Possible Groupings
We need to divide 7 people into 3 groups (for 3 vehicles) such that each group has at least 1 person, and no group has more than 5 people.

### Step 2: Partitioning the 7 People
Since we need to ensure that all three vehicles are used, each group must have at least 1 person. This converts our problem into distributing the 7 people into three groups with non-zero occupants, where no group exceeds 5 people.

We are essentially looking for all the solutions to the equation:

\[
x_1 + x_2 + x_3 = 7
\]

where:
- \(x_1, x_2, x_3 \geq 1\) (each vehicle has at least one person)
- \(x_1, x_2, x_3 \leq 5\) (each vehicle can hold at most 5 people)

### Step 3: Transform the Equation
Let’s substitute \(x_i' = x_i - 1\) (so that \(x_i' \geq 0\)), which gives us:

\[
(x_1' + 1) + (x_2' + 1) + (x_3' + 1) = 7
\]

This simplifies to:

\[
x_1' + x_2' + x_3' = 4
\]

Now, we need to count the non-negative integer solutions to this equation with the constraint that \(x_i' + 1 \leq 5\), or equivalently, \(x_i' \leq 4\).

### Step 4: Find the Number of Solutions
We now find how many ways there are to distribute 4 units (the "extra" people) among 3 groups, where each group can get between 0 and 4 people.

Using the stars and bars method for unrestricted distributions, the number of ways to distribute 4 people into 3 groups is:

\[
\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15
\]

### Step 5: Factor in All Three Vehicles
Each distribution can be assigned to any of the three vehicles, and we need to consider the possible permutations of the groups, meaning that the total number of ways to assign these groups to the vehicles (since each vehicle must be used) is \(3! = 6\).

### Step 6: Final Answer
Thus, the total number of ways the 7 people can visit such that all three vehicles are used is:

\[
15 \times 6 = 90
\]

### Conclusion:
There are **90 ways** in which the 7 people can visit the marriage function using all three vehicles, with each vehicle being used and accommodating between 1 and 5 people.

---

Do you want more details or have any questions?

Here are some related questions:
1. How do you apply stars and bars to more complex partitions?
2. What is the inclusion-exclusion principle in combinatorics?
3. How would the solution change if the vehicle capacities were different?
4. Can you solve this problem if only two vehicles were used instead of three?
5. What happens if the number of people and vehicles change?

**Tip:** The stars and bars method is a powerful tool in combinatorics for solving distribution problems where the order doesn't matter but constraints on group sizes might apply.

Seven people visited a marriage function in three different vehicles, each accommodating a maximum of 5 people. How many ways can they visit such that they use all three vehicles?

Explore how to solve a combinatorics problem using the stars and bars method and inclusion-exclusion principle. Learn how to distribute 7 people across 3 vehicles with a capacity constraint of 5 per vehicle. Suitable for Grades 11-12 and undergraduate math students.

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