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Nov 2005 p6 q3
2809
A staff car park at a school has 13 parking spaces in a row. There are 9 cars to be parked.
How many different arrangements are there for parking the 9 cars and leaving 4 empty spaces? [2]
How many different arrangements are there if the 4 empty spaces are next to each other? [3]
If the parking is random, find the probability that there will not be 4 empty spaces next to each other. [2]
Solution
(i) To find the number of different arrangements for parking the 9 cars and leaving 4 empty spaces, we use permutations. The total number of ways to arrange 9 cars in 13 spaces is given by the permutation formula:
(ii) If the 4 empty spaces are next to each other, treat them as a single block. This reduces the problem to arranging 10 blocks (9 cars + 1 block of empty spaces). The number of arrangements is:
\(10! = 3,628,800\).
(iii) The probability that there will not be 4 empty spaces next to each other is the complement of the probability that they are next to each other. Thus,
\(P(\text{not next to each other}) = 1 - \frac{\text{arrangements with 4 empty spaces together}}{\text{total arrangements}}\).
