I was given a homework question that reads:
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
My reasoning:
There are 6! possibilities. there are 3! ways of re-arranging the 3 cities that are neither denver nor san francisco.
We can see the following patterns
$D_1,S,\_,\_,\_,\_$
$D_1,D_2,S,\_,\_,\_$
$D_1,D_2,D_3,S,\_,\_$
$D_1,D_2,D_3,D_4,S,\_$
$D_1,D_2,D_3,D_4,D_5,S$
Where $D_i$ represents a possible position for denver.
There are thus 1 combinations in teh first case, 2 in the second...
Giving us a total of 3!$(1+2+3+4+5)=90$ possibilities for travelling to denver before san francisco.
90/6!=0.125
My professor's answer
Denver - San Francisco - one of six cities) and (San Francisco - Denver - one of six cities); hence, the probability that she will visit Denver before San Fransico is 0.5.
Who is correct, and if I am wrong, why is my reasoning wrong?
There are $4$ cities that are neither Denver nor San Francisco. Your method works fine, but replacing $3!$ by $4!$ gives $4\cdot0.125=0.5$ as the correct solution, which is what your professor decided as well.