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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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up vote 1 down vote accepted

The itinerary is randomly chosen, so all itineraries are equally likely. Thus, the probability of visiting in alphabetical order is the number of alphabetically ordered trips divided by the total number of possible trips.

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 $\frac{1}{12!}$.

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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.

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