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June 2023 p52 q6
2789
For another competition, a team of 9 people consists of 2 swimmers, 3 cyclists, and 4 runners. The team members stand in a line for a photograph.
(b) How many different arrangements are there of the 9 people if the swimmers stand together, the cyclists stand together, and the runners stand together?
(c) How many different arrangements are there of the 9 people if none of the cyclists stand next to each other?
Solution
(b) Treat each group (swimmers, cyclists, runners) as a single unit. This gives us 3 units to arrange, which can be done in \(3!\) ways. Within each group, the swimmers can be arranged in \(2!\) ways, the cyclists in \(3!\) ways, and the runners in \(4!\) ways. Therefore, the total number of arrangements is \(3! \times 2! \times 3! \times 4! = 6 \times 2 \times 6 \times 24 = 1728\).
(c) First, arrange the 6 non-cyclists (2 swimmers and 4 runners) in \(6!\) ways. This creates 7 gaps (before, between, and after these 6 people) where the 3 cyclists can be placed. Choose 3 out of these 7 gaps to place the cyclists, which can be done in \(\binom{7}{3}\) ways. The cyclists themselves can be arranged in \(3!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).
