There are 15 European cities that Brian would eventually like to visit. On his next vacation, though, he only has time to visit 3 of the cities: one on Monday, one on Tuesday, and one on Wednesday. He is now trying to make a schedule of which city he'll visit on which day. How many different schedules are possible? (Assume that he will not visit a city more than once.)
Question
There are 15 European cities that Brian would eventually like to visit. On his next vacation, though, he only has time to visit 3 of the cities: one on Monday, one on Tuesday, and one on Wednesday. He is now trying to make a schedule of which city he'll visit on which day. How many different schedules are possible? (Assume that he will not visit a city more than once.)
Solution
To solve this problem, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order. The order of arrangement of the object is very important. The number of permutations of n objects taken r at a time is given by:
P(n, r) = n! / (n-r)!
where n! denotes the factorial of n.
In this case, Brian has 15 cities to choose from and he wants to visit 3 of them. So, n = 15 and r = 3.
Let's calculate the number of permutations:
P(15, 3) = 15! / (15-3)!
= 151413121110987654321 / 12111098765432*1
After canceling out the common factors, we get:
= 151413
= 2730
So, there are 2730 different schedules possible for Brian's trip.
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