Statistics- Counting problem and probability?

Question

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 the route.

a) how many different itineraries ( and trip costs) are possible)?

Since order is significant I used the equation for permutations and arrived at an answer of 720 possible itineraries. (I hope this much is correct)

b) If the business woman selects one of the possible itineraries and Denver and San Francisco are two of the cities she plans to visit, what is the probability she will visit Denver before San Francisco?

What I think I have to do here is figure out:

p(visit Denver before SF) = (# itineraries where Denver is before SF) / (total # itineraries)

My problem is figuring out how many itineraries have Denver before SF, assuming what I have done is correct up to this point. I'll award points to the best answer. Thanks.

Answer 1

a) Correct

b) P(visit Denver before SF)=(# itineraries where Denver is before SF) / (total # itineraries)=m/n, where n=6!=720

and m=4!*[(1/2)*(5V2)+5], where 5V2=5!/(5-2)!=5*4 is the number of variations without repeat of 2 elements out of 5 ==>

m=24*[(1/2)*20+5]=360 ==>

P=360/720=1/2

Why m=4!*[(1/2)*(5V2)+5] ?

Without Denver and San Francisco there are 4 cities left, which could be arranged in 4! different ways.

Denver and San Francisco could be visited:

1) in 5 different ways with SF following directly after Denver ("in a package"):

(D,SF),_,_,_,_

_,(D,SF),_,_,_

_,_,(D,SF),_,_

_,_,_,(D,SF),_

_,_,_,_,(D,SF)

2) in (1/2)*(5V2)=10 different ways separated (Denver before SF).

I hope you could figure it out. If not, don't hesitate to ask.