A team of 8 students is going on an excursion using two cars. One car has 5 seats and the other car has 4 seats. In how many different ways can the 8 students be divided between the two cars, assuming seats within a car are not distinguished?

Introduction / Context:
This question involves distributing people into groups with capacity constraints. Instead of focusing on arranging people in specific seats, we only care about which students go in which car. Such distribution problems come up frequently in combinatorics and in real life planning scenarios. Understanding how to split a group into subgroups according to given size limits is a key counting skill.

Given Data / Assumptions:

• Total number of students: 8.
• Car 1 has 5 seats.
• Car 2 has 4 seats.
• There will be exactly 8 students travelling, so one seat will remain empty.
• Students are distinct individuals.
• We do not distinguish individual seats inside a car, only which students are in each car.

Concept / Approach:
We must count all valid ways of dividing 8 distinct students between two cars with capacities 5 and 4. Let the number of students in the 5 seat car be x and in the 4 seat car be y. We have two constraints:

• x + y = 8 (all students travel).
• 0 ≤ x ≤ 5 and 0 ≤ y ≤ 4 (capacity limits).

Solving these, only two valid distributions exist: (x, y) = (4, 4) and (5, 3). We then count combinations for each and add them.

Step-by-Step Solution:
Step 1: Determine possible splits: x + y = 8 with x ≤ 5 and y ≤ 4. Step 2: Valid splits are (4, 4) and (5, 3). No other combination fits both constraints. Step 3: For (4, 4): choose 4 students from 8 for the first car. Number of ways = 8C4. Step 4: Compute 8C4 = 70. The remaining 4 automatically go in the second car. Step 5: For (5, 3): choose 5 students from 8 for the 5 seat car. Number of ways = 8C5. Step 6: Compute 8C5 = 56. The remaining 3 go in the 4 seat car. Step 7: Total ways = 70 + 56 = 126.

Verification / Alternative check:
We can see that 8C4 = 8C4 and 8C5 = 8C3 by symmetry of combinations. So the total 8C4 + 8C3 = 70 + 56 is consistent with the counting of different ways to pick which students use which car under the given capacity constraints. There is no double counting, since each split has a unique number of students in each car.

Why Other Options Are Wrong:
240: Might come from multiplying 8C4 by 2 incorrectly or including invalid distributions. 120: Close to 5!, but this does not match the exact counting logic here. 260: A random sum that does not align with any correct combination breakdown. Only 126 correctly matches the sum of valid distributions (4, 4) and (5, 3).

Common Pitfalls:
Some students ignore car capacities and assume any split is acceptable, which leads to overcounting. Others mistakenly treat the two cars as indistinguishable, which would require dividing by 2 in some symmetric cases, but here the cars have different capacities, so they are inherently distinct. Finally, it is important not to attempt to assign specific seats inside each car, as the problem only asks for distributions by car, not seat level arrangements.

Final Answer:
The number of different ways in which the 8 students can travel in the two cars is 126.