Question

Probability. I'm not seeing how to get b).

A businesswoman in Philadelphia is preparing an itinerary for a visit to 6 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? \(6!\)=720

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 D before SF?

D=denver, S=San Francisco

Answer 1

this is to find the number of itineraries where D is before S.

\(\underbrace{D S}\)_ _ _ _ = 5!
1 2 3 4 5

D _ S _ _ _ = 4!
D _ _ S_ _ =3!
D _ _ _ S _ = 2!
D _ _ _ _ S =1

and then adding them up but it's not 360 (what the answer apparently is)

What am i missing?

Answer 2

for (b) by symmetry the answer is 1/2, but if you want to count.

\[4!5+4!4+4!3+4!2+4!\]
\[4!(5+4+3+2+1)=24\cdot15=360\]

Answer 3

ohh i wasn't taking the identities of the blanks into account, thanks so much for your help.