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Can anyone help with this? Can't wrap my head around this problem.
A staff car park at a school has 13 parking spaces in a row. There are 9 cars to be parked.
i. How many Different arrangements are there for parking the 9 cars and leaving 4 empty spaces?
ii.How many different arrangements are there if the 4 empty spaces are next to each other?
iii. if the parking is random, find the probability that there will not be 4 empty spaces next to each other.
Any help would be appreciated.
I'll assume the 9 cars are all different, so that permutation (rather than combination) is used:
(i)
Choose 9 slots from 13, order is considered.
13P9 = (13!)/(4!) = 259459200
(ii)
Now we need to make sure that in every case, the 4 empty slots are bound together.
We can group the 4 empty slots as a single unit, so that:
- there are 9 units to park cars
- there is a single unit of 4 empty slots
A total of 10 units.
Thus the case is to choose 9 units from 10, order is considered.
10P9 = (10!)/(1!) = 3628800
(iii)
The total number of possible outcomes is the answer to (i), which is 259459200.
The number of possible outcomes with 4 empty spaces next to each other, is the answer to (ii), 3628800.
So the number of outcomes where the 4 empty spaces are not next to each other = 259459200 - 3628800.
Probability = (259459200 - 3628800) / 259459200 = 0.986