Twenty distinct cars park in the same parking lot every day. Ten of these cars a

Question

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. (a) In how many different ways can the cars line up? (b) 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)?

Explanation / Answer

a. Each car can take one position. So, number of positions for 1st car = 20, for 2nd car = 19 and so on for last car =1

Hence the required number of ways in which they can line up = 20*19*18*..*1 = 20!

b. First car has 20 positions. Next car has 9 or 10 positions, depending on if it is same as 1st or not. Next car may have 10 or 9 or 8 positions. SImilarly, numnber of ways in which cars can park in this manner = 20 * 10*9*..1 * 9*8*..*1 = 20*10!*9!

Hence required probability = 20*10!*9!/20! = 10!*9!/19! = 1/ 19C10

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