7 people visited a marriage function in 3 different vehicles each with maximum 5 people how many ways they can use such that they all uses 3 vehicles

We are given that 7 people are visiting a marriage function in 3 different vehicles, each with a capacity of 5 people. The goal is to find how many ways they can distribute themselves across these vehicles such that all 7 people use the 3 vehicles.

### Step 1: Basic understanding of the problem

- Total number of people: 7
- Total number of vehicles: 3
- Maximum capacity per vehicle: 5 (but we won't exceed this as only 7 people are involved).

We need to partition 7 people into 3 groups where none of the groups are empty (since all vehicles must be used). 

### Step 2: Possible partitions of 7 people across 3 vehicles

Since each vehicle must carry at least one person, the ways we can partition 7 people into 3 vehicles can be expressed as solutions to the equation:
\[
x_1 + x_2 + x_3 = 7
\]
where \(x_1, x_2, x_3 \geq 1\) (i.e., each vehicle carries at least one person). To simplify, let’s redefine the equation as:
\[
y_1 + y_2 + y_3 = 4
\]
where \(y_1 = x_1 - 1\), \(y_2 = x_2 - 1\), and \(y_3 = x_3 - 1\), meaning each \(y_i \geq 0\). Now, we are looking for the number of non-negative integer solutions to this equation.

This is a well-known problem of finding the number of non-negative integer solutions to an equation, which is given by the stars and bars formula:
\[
\text{Number of solutions} = \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15
\]
So, there are 15 ways to partition the 7 people into 3 non-empty groups.

### Step 3: Permuting people within groups

Once the group sizes are fixed, the next step is to assign specific people to the groups. Suppose the group sizes are \(a\), \(b\), and \(c\) such that \(a + b + c = 7\), with \(a \geq b \geq c\). For each distinct partition, the number of ways to assign people to the vehicles is the multinomial coefficient:
\[
\frac{7!}{a!b!c!}
\]

For each possible partition, we compute the corresponding multinomial coefficient, and then we sum over all possible partitions.

### Step 4: Total number of ways

Let's calculate the number of ways based on different partitions of the number 7 into 3 parts:

1. **Partition (5, 1, 1)**:
   \[
   \frac{7!}{5!1!1!} = \frac{5040}{120 \times 1 \times 1} = 42
   \]
2. **Partition (4, 2, 1)**:
   \[
   \frac{7!}{4!2!1!} = \frac{5040}{24 \times 2 \times 1} = 105
   \]
3. **Partition (3, 3, 1)**:
   \[
   \frac{7!}{3!3!1!} = \frac{5040}{6 \times 6 \times 1} = 140
   \]
4. **Partition (3, 2, 2)**:
   \[
   \frac{7!}{3!2!2!} = \frac{5040}{6 \times 2 \times 2} = 210
   \]

Summing these values:
\[
42 + 105 + 140 + 210 = 497
\]

Thus, there are 497 distinct ways for the 7 people to use the 3 vehicles such that each vehicle is used.

---

Would you like further details or clarification on any part of the solution? Here are some related questions to consider:

1. How would the solution change if one vehicle could remain empty?
2. What if there were 8 people and the same vehicle constraints?
3. How would the answer change if two vehicles could accommodate up to 6 people?
4. What is the logic behind the stars and bars method?
5. Could this problem be solved using another combinatorial technique?

**Tip**: The stars and bars method is a great tool for dividing groups or items into distinct categories. It helps solve problems involving the partitioning of objects.

7 people visited a marriage function in 3 different vehicles, each with a maximum of 5 people. How many ways can they distribute themselves such that all of them use the 3 vehicles?

Discover how to calculate the number of ways to distribute 7 people across 3 vehicles using combinatorics. This problem involves partitioning 7 people into groups where all vehicles must be used, utilizing the stars and bars method and multinomial coefficients.

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