7 people visited a marriage Function in   3 dier. ent vehicles, each accommodating a maximum of 5 persons. In how many ways can they visit, such that they use all the 3 vehicles

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**.

---

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.

7 people visited a marriage function in 3 different vehicles, each accommodating a maximum of 5 persons. In how many ways can they visit, such that they use all the 3 vehicles?

Discover how to solve a combinatorics problem involving 7 people traveling in 3 vehicles, with each vehicle holding up to 5 persons. This problem utilizes combinations and permutations to partition groups and assign vehicles, making it perfect for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation