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

We are given that 7 people are visiting a marriage function in 3 different vehicles, with each vehicle having a maximum capacity of 5 persons. We need to find the number of ways they can travel such that all 3 vehicles are used.

### Step 1: Split the 7 people among the 3 vehicles
To use all 3 vehicles, we must distribute the 7 people such that no vehicle is empty. The possible ways to distribute 7 people among the 3 vehicles (with no vehicle being empty) are:

- 5 people in one vehicle, 1 person in another, and 1 person in the third.
- 4 people in one vehicle, 2 people in another, and 1 person in the third.
  
### Step 2: Number of ways to assign people to the vehicles
For each of these distributions, we calculate the number of ways to assign the 7 people:

1. **Case 1**: 5 people, 1 person, 1 person.
   - Choose 5 people for the first vehicle from the 7 people:  
     \[
     \binom{7}{5} = \binom{7}{2} = 21
     \]
   - Choose 1 person for the second vehicle from the remaining 2 people:  
     \[
     \binom{2}{1} = 2
     \]
   - The last remaining person will go in the third vehicle:  
     \[
     1 \text{ way}
     \]
   Therefore, the total number of ways for this case is:  
     \[
     21 \times 2 \times 1 = 42 \text{ ways}
     \]

2. **Case 2**: 4 people, 2 people, 1 person.
   - Choose 4 people for the first vehicle from the 7 people:  
     \[
     \binom{7}{4} = 35
     \]
   - Choose 2 people for the second vehicle from the remaining 3 people:  
     \[
     \binom{3}{2} = 3
     \]
   - The last remaining person will go in the third vehicle:  
     \[
     1 \text{ way}
     \]
   Therefore, the total number of ways for this case is:  
     \[
     35 \times 3 \times 1 = 105 \text{ ways}
     \]

### Step 3: Permutations of the vehicles
Since the vehicles are distinguishable, we must account for the different ways to assign the groups to the 3 vehicles. For each case, we multiply by the number of ways to permute 3 vehicles, which is \( 3! = 6 \).

1. For the first case (5, 1, 1), the number of ways is:
   \[
   42 \times 6 = 252 \text{ ways}
   \]

2. For the second case (4, 2, 1), the number of ways is:
   \[
   105 \times 6 = 630 \text{ ways}
   \]

### Step 4: Total number of ways
The total number of ways is the sum of the two cases:
\[
252 + 630 = 882 \text{ ways}
\]

Thus, the total number of ways they can visit such that they use all 3 vehicles is \( \boxed{882} \).

Would you like more details on any step, or do you have any questions?

Here are some related questions:
1. What if the vehicles were identical? How would that change the answer?
2. How does this problem relate to the stars and bars method in combinatorics?
3. What if the vehicles had different capacities?
4. How would the answer change if the 7 people included 2 VIPs that must sit alone?
5. How do you calculate the number of ways if no vehicle had a fixed capacity limit?

**Tip**: When assigning people to groups with specific size limits, breaking the problem into cases can simplify complex distributions.

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 3 vehicles?

Explore how to distribute 7 people across 3 vehicles with a maximum capacity of 5 persons each, ensuring all vehicles are used. This combinatorics problem involves binomial coefficients and permutation calculations. Perfect for high school students in Grades 10-12.

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