7 people visited a marriage in 3 diff vehicle, each accommodating a max of 5 person. In how many ways they can visit such that they use all 3 vehicle 

The problem can be solved using combinatorics. Here's the detailed approach:

### Step 1: Assign people to the vehicles
We need to assign 7 people to 3 vehicles, with each vehicle accommodating a maximum of 5 people. Since all three vehicles must be used, each vehicle should carry at least 1 person. 

Thus, we need to find the number of ways to distribute 7 people into 3 groups (vehicles) where none of the groups is empty, and each group has no more than 5 people.

### Step 2: Determine possible group sizes
Let the number of people in the three vehicles be \( x_1, x_2, x_3 \), where:
\[
x_1 + x_2 + x_3 = 7
\]
and \( 1 \leq x_1, x_2, x_3 \leq 5 \).

The possible partitions of 7 people into 3 non-empty groups, where no group has more than 5 people, are:
- \( (5, 1, 1) \)
- \( (4, 2, 1) \)
- \( (3, 3, 1) \)
- \( (3, 2, 2) \)

### Step 3: Count the arrangements for each case
#### Case 1: \( (5, 1, 1) \)
- Choose 5 people for the first vehicle: \( \binom{7}{5} = 21 \)
- Choose 1 person for the second vehicle: \( \binom{2}{1} = 2 \) (as 2 people remain after choosing 5)
- The remaining person goes to the third vehicle automatically: 1 way
- Total for this case: \( 21 \times 2 = 42 \)

#### Case 2: \( (4, 2, 1) \)
- Choose 4 people for the first vehicle: \( \binom{7}{4} = 35 \)
- Choose 2 people for the second vehicle: \( \binom{3}{2} = 3 \)
- The remaining person goes to the third vehicle: 1 way
- Total for this case: \( 35 \times 3 = 105 \)

#### Case 3: \( (3, 3, 1) \)
- Choose 3 people for the first vehicle: \( \binom{7}{3} = 35 \)
- Choose 3 people for the second vehicle: \( \binom{4}{3} = 4 \)
- The remaining person goes to the third vehicle: 1 way
- Total for this case: \( 35 \times 4 = 140 \)

#### Case 4: \( (3, 2, 2) \)
- Choose 3 people for the first vehicle: \( \binom{7}{3} = 35 \)
- Choose 2 people for the second vehicle: \( \binom{4}{2} = 6 \)
- The remaining 2 people go to the third vehicle automatically: 1 way
- Since the two groups of 2 people are indistinguishable, we must divide by 2 to account for overcounting:
- Total for this case: \( \frac{35 \times 6}{2} = 105 \)

### Step 4: Add the results
The total number of ways to assign the 7 people to 3 vehicles such that all vehicles are used is:
\[
42 + 105 + 140 + 105 = 392
\]

Thus, the total number of ways they can visit such that all 3 vehicles are used is **392**.

Would you like further details or have any questions?

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Here are 5 related questions you can explore:
1. How does the number of ways change if each vehicle can accommodate a maximum of 6 people?
2. What if there are 8 people and the same vehicle capacity constraints?
3. How would the solution change if only 2 vehicles are available, each still accommodating up to 5 people?
4. What if each vehicle must carry at least 2 people?
5. How many ways are there to arrange 7 people into 3 groups if there is no limit on group size?

**Tip:** When working with combinatorics problems, always pay attention to the conditions (e.g., minimum group size, overcounting) to avoid errors.

7 people visited a marriage in 3 diff vehicle, each accommodating a max of 5 person. In how many ways they can visit such that they use all 3 vehicle?

Explore how to distribute 7 people across 3 vehicles, each with a maximum capacity of 5 people, ensuring all vehicles are used. This combinatorics problem involves partitioning and calculating combinations. Suitable for high school students in Grades 10-12.

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