A team of 8 students goes on an excursion, in two cars, of which one can seat…

Concept: When a fixed group of n distinct people must be split between two vehicles/positions of given sizes, the number of ways for one particular split is nCr (choose which r of the n take one vehicle; the rest automatically take the other). If more than one split of sizes is capable of accommodating everyone, the splits are mutually exclusive outcomes of the same trip, so their counts are added, not multiplied, to get the total.

Application: Apply this to the 8 students and the two cars (capacities 5 and 4, total capacity 9, so exactly one seat overall stays empty):

1. Since only 8 of the 9 total seats are filled, there are exactly two ways the empty seat can arise: the 5-seat car is completely full (5 students there, 3 in the 4-seat car), or the 4-seat car is completely full (4 students there, 4 in the 5-seat car).

2. Case I — 5-seat car full: choose which 5 of the 8 students ride in it; the remaining 3 automatically ride in the other car. Number of ways = 8C5 = 8!/(5!·3!) = 56.

3. Case II — 4-seat car full via the larger car: choose which 4 of the 8 students ride in the 5-seat car; the remaining 4 automatically ride in the 4-seat car (which is then exactly full). Number of ways = 8C4 = 8!/(4!·4!) = 70.

4. A given group of 8 travelling companions falls into exactly one of these two cases, never both at once, so the two case-counts are added together to get the total number of ways to travel: 56 + 70 = 126.

Cross-check: Look at it from the 4-seat car's side instead: since the 5-seat car can hold at most 5, the 4-seat car must carry at least 3 of the 8 students, so it carries either 3 or 4. A 3-in-the-4-seat-car split is the same grouping as “5 in the 5-seat car,” and a 4-in-the-4-seat-car split is the same as “4 in the 5-seat car.” The same two cases reappear from this direction and, added together, reproduce the same total — confirming the count.