Everyday six visitors enter a residential complex and park their cars in one of the 10 adjacent parking spaces. The three lawyers among these five visitors arrive earlier than the other 3 visitors and never park in adjacent slots. Find the number of different ways in which the six cars can be arranged in the 10 spaces (linearly) such that there are no empty slots between any of the cars.
Explanation:
Let us first calculate the number of ways of arranging 6 cars linearly such that the three lawyers are never together.
3 cars of non-lawyers can be arranged in 3! = 6 ways.
| C | C | C |
Now, the cars of lawyers can be placed in between these 3 cars in 4C3 × 3! = 24 ways
∴ Number of ways of arranging cars = 6 × 24 = 144 ways.
Parking slots ⇒ 1 2 3 4 5 6 7 8 9 10
Number of ways of selecting 6 adjacent parking slots = 5
∴ Total number of ways of parking these 6 cards in 6 adjacent slots = 5 × 144 = 720 ways.
Hence, option (a).
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