A team of 8 students is going on an excursion in two cars. One car has a seating capacity of 5 and the other car has a seating capacity of 4. Since there are only 8 students and 9 seats, one seat will remain empty. In how many different ways can the 8 students travel in the two cars?

Introduction / Context:
This question tests the understanding of distributing people into distinct groups with capacity constraints. We must assign 8 students to two cars of capacities 5 and 4, with one seat necessarily left vacant, and count the total number of valid distributions. This is a combinatorial distribution problem frequently seen in reasoning and aptitude tests.

Given Data / Assumptions:

• There are 8 distinct students.
• Car A has 5 seats and Car B has 4 seats.
• Total seats available: 9.
• Since there are only 8 students, exactly one seat will be empty.
• We are counting distinct ways of assigning students to cars, ignoring the specific seat positions within each car.

Concept / Approach:
We consider how many students go in each car under the seating constraints. There are two possible occupancy patterns consistent with 8 students and car capacities 5 and 4: Car A has 5 students and Car B has 3 students, or Car A has 4 students and Car B has 4 students. For each case, we count how many subsets of students fit the described distribution and sum the counts because the cases are mutually exclusive.

Step-by-Step Solution:
Step 1: Case 1: Car A carries 5 students, Car B carries 3 students (one vacant seat in Car B). Choose 5 students out of 8 for Car A.Step 2: The number of ways to choose 5 students for Car A is 8C5 = 56. The remaining 3 automatically go to Car B.Step 3: Case 2: Car A carries 4 students, Car B carries 4 students (one vacant seat in Car A). Choose 4 students out of 8 for Car B or Car A, the count is symmetric.Step 4: The number of ways to choose 4 students for Car B is 8C4 = 70. The remaining 4 go to Car A.Step 5: The two cases are disjoint, so total assignments = 56 + 70 = 126.Step 6: Therefore, there are 126 different ways for the students to travel in the two cars.
Verification / Alternative check:
We can also think in terms of choosing which seat remains empty, but since we are not distinguishing individual seats within the cars, this would overcomplicate the count. The combinational approach based on how many students ride in each car is simpler and directly uses the capacities to find all possibilities. The count 126 lies between 2^8 and 3^8 and seems reasonable for two groupings with capacity constraints.

Why Other Options Are Wrong:
120 is very close but would correspond to a different counting structure, such as ignoring one of the valid occupancy patterns. Values like 146 or 156 do not arise from any natural combination of the two case counts. Only 126 exactly equals 8C5 + 8C4 and correctly accounts for both possible distributions of students between the two cars.

Common Pitfalls:
Students sometimes assume that Car A must always be full, ignoring the possibility that Car B could be full instead. Another common error is to consider 8C5 only, leading to 56 and missing the second case entirely. Recognising that one seat can be empty in either car helps identify both valid patterns.

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