# A tourist in France wants to visit 12 different cities, find the probability.

The Question :

A tourist in France wants to visit 12 different cities. If the route is randomly selected, what is the probability that she will visit the cities in alphabetical order?

I thought about this in two ways and I got two different answers, both of them look correct to me ! but only one is correct.

1/144 (12 cities times 12 times to make the arrangement)

1/479001600 (which is generated by the permutation 12 P 12)

Which one is correct? and is there a better answer?

Please explain how do you know if the problem is permutation or combination or just simple multiplication.

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The order you visit the cities matters, so the total number of itineraries is $P(12,12) = 12!$. Since only one of these is in alphabetical order, the probability of randomly choosing alphabetical order is $\frac{1}{12!}$.
Your $\frac 1{12!}$ is correct. There is $\frac 1{12}$ chance that the first city is correct. Assuming it is, there is $\frac 1{11}$ that the next one is correct, and so on. Multiplying all these gives the answer.