(iii) Treat the 3 buses as a single unit. This gives us 7 units to arrange: 6 cars and 1 bus unit. The number of ways to arrange these 7 units is given by:
\(7!\)
Within the bus unit, the 3 buses can be arranged among themselves in:
\(3!\)
Thus, the total number of arrangements is:
\(7! \times 3! = 5040 \times 6 = 30240\)
(iv) Place a car at each end, leaving 4 cars and 3 buses to arrange in the middle. The number of ways to arrange 4 cars and 3 buses such that no two buses are adjacent is equivalent to arranging the 4 cars first and then placing the buses in the gaps between and around the cars. There are 5 gaps (before, between, and after the cars) to place the buses:
Choose 3 out of these 5 gaps to place the buses:
\(\binom{5}{3} = 10\)
Arrange the 4 cars:
\(4!\)
Arrange the 3 buses:
\(3!\)
Thus, the total number of arrangements is:
\(4! \times \binom{5}{3} \times 3! = 24 \times 10 \times 6 = 1440\)
Multiply by 6 for the 6 possible cars that can be at each end:
\(6 \times 1440 = 8640\)
However, the mark scheme indicates a different approach, leading to:
\(6! \times 5P3 = 43200\)
