The problem in the image states:

"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?"

Solution:

1. Understanding the distribution problem:

   • We need to distribute 7 people into 3 vehicles.
   • Each vehicle must carry at least 1 person, and no vehicle can carry more than 5 people.

2. Step 1: Breaking down the number of people into vehicles. Since there are 7 people and all vehicles must be used, we need to find how to split 7 people into groups where each group has at least 1 person.

   This is a classic combinatorics problem of distributing people into groups with restrictions. The combination of groups is solved using the stars and bars approach and factorial calculation for permutations.

3. Step 2: Distribution options The only valid group distributions are:

   • (3, 2, 2): One vehicle has 3 people, and the other two vehicles have 2 people each.

4. Step 3: Calculating the number of ways to assign people:

   • First, choose 3 people to go in the vehicle carrying 3 people. This can be done in $\binom{7}{3} = \frac{7!}{3!(7-3)!} = 35$ ways.
   • After assigning 3 people, we have 4 people left. We need to assign 2 people to one vehicle and 2 to the other.
   • The number of ways to split these 4 people into two groups of 2 is $\frac{\binom{4}{2}}{2!}$. We divide by $2!$ because the two groups of 2 are indistinguishable. This gives $\frac{6}{2} = 3$ ways.

5. Step 4: Total number of ways: The total number of ways to distribute the 7 people into the 3 vehicles is $35 \times 3 = 105$.

However, upon review, we need to adjust for ordering of vehicles and revisit combinatorics for constraints,