Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made. The parking lot has exactly twenty spaces, all in a row. so the cars park side by side. However, the drivers have varying schedules, so the position any car might take on a certain day is random. In how many different ways can the cars line up? What is the probability that on a given day, the cars will park in such a way that they alternate (no two US-made are adjacent and no two foreign-made are adjacent) ?

Question

Skyler Carlson

Engineering

15 January, 16:02

Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made. The parking lot has exactly twenty spaces, all in a row. so the cars park side by side. However, the drivers have varying schedules, so the position any car might take on a certain day is random. In how many different ways can the cars line up? What is the probability that on a given day, the cars will park in such a way that they alternate (no two US-made are adjacent and no two foreign-made are adjacent) ?

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Answers (1)

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Angelo Duran · Answer 1

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random. Therefore, number ways in which cars can line up is given as:

No. of ways = 20! = 2.43 x 10^18

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!) / 20!

Probability = 0.00001 = 0.001%