Probability of visiting a city before another A businesswoman in Philadelphia is preparing an itinerary for a visit to six major cities. The distance traveled, and hence the cost of the trip, will depend on the order in which she plans her route. 
a) How many different itineraries (and trip costs) are possible? 
b) If the businesswoman randomly selects one of the possible itineraries and Denver and San Francisco are two of the cities that she plans to visit, what is the probability that she will visit Denver before San Francisco? 
I think I understand that the number of itineraries is simply 6! because there are six cities and each time we select one we have to take that into account.
But for the second part, I am not sure what to do, does it start with findingthe probability of visiting Denver and San Francisco?
Any help or hint is well appreciated, thank you in advance!
 A: Hint:  presumably she visits all six cities and the only question is what order.  For every order that has Denver before San Francisco, the opposite order has San Francisco before Denver.
A: How convenient that this question was asked less than a week ago. I am currently working through old statistics problems because I am trying to prepare for the SOA - P exam, and this is the question I was just working on.
Nevertheless, here was my solution to the problem.


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*a) It is apparent that order will matter here since if our cities are $A, B, C, D, E,$ and $F$, the itinerary $ABCDEF$ is different from itinerary $ABCDFE$. (Think about it using a connect-the-dots analogy. The shape you make will be completely different if you don't connect the dots in the right order.) Since order does matter, we will need to use a permutation. So, we have $6$ cities, and we're going to be permuting all $6$ of them. Thus there are $$N = P^6_6 = {6!\over(6-6)!} = {6!\over0!} = 6! = 720$$ possible itineraries. That is, the sample space, $S$, for possible itineraries has $720$ sample points

*b) We first need to determine the number of sample points in the sample space for the experiment $T$, the sample points in $S$ such that Denver, $D$, comes before San Francisco, $F$. Here, we're going to be fixing two cities to be constant. Hence, we will have $6$ cities, but permuting $4$ of them. In other words, $$N_T = P^6_4 = {6!\over(6-4)!} = {6!\over 2!}.$$ Thus, we use this result to give us $$P(\text{visit $D$ before $F$}) = {N_T\over N} = {6!\over 2!}\cdot{1\over 6!} = {1\over2!} = {1\over 2} = 0.5.$$

