Bunuel wrote:

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

A. 120

B. 126

C. 146

D. 156

E. 166

We can call the 5-seat car Car A and the 4-seat car Car B. Therefore, we have two cases to consider:

Case 1: Car A seats 5 and Car B seats 3

Case 2: Car A seats 4 and Car B seats 4

Case 1:

If Car A seats 5 and Car B seats 3, then we have 8C5 x 3C3 = (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2) x 1 = 8 x 7 = 56 ways for the 8 students to travel in these two cars.

Case 2:

If Car A seats 4 and Car B seats 4, then we have 8C4 x 4C4 = (8 x 7 x 6 x 5)/(4 x 3 x 2) x 1 = 2 x 7 x 5 = 70 ways for the 8 students to travel in these two cars.

So we have a total of 56 + 70 = 126 ways for the 8 students to travel in these two cars.

Answer: B.

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