Stats question?

Question

A business woman is preparing an itinerary for a vist to six major cities

a) How many different itineraries are possible?
which is easy, 6!

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 Denver before San Francisco.

I have the answer, I just did [(5! x 1) + (4 x 4!) + (12 x 3!) + (24 x 2!) + (4!)] / 6!

Now my question is, is there a more elegant way to extract the same answer?

Answer 1

It is obvious without any calculation that the probability is 0.5.

(The position of Denver and San Francisco in this question is absolutely symmetrical.)
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Answer 2

Probability of the woman starting in Denver is 1/6. Probability of her starting the trip in SF is also 1/6. That leaves us with a probability of 4/6 that the trip started neither in Denver or SF. In that case, the probability of Denver being the second city is 1/5, and the probability of the second city not being either Denver or SF 3/5.... which boils down to:

1/6 + 4/6( 1/5 + 3/5( 1/4 + 2/4( 1/3 + 1/3 * 1/2 ))) = 1/2

Answer 3

(4! + 6*(3! + 5*(2! + 4)) / 6! = 240 / 720 = 1/3