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?

A) 720 possible itineraries

B) 0.5

Step-by-step explanation:

She visited 6 cities.

A)Now, to get the number of different itineraries that are possible, we will use n! = n(n-1)(n-2)(n-3)

In this question, n is 6

Thus;

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

B) First of all, let's calculate the number of sample points in the sample space for the number of cities visited. Since we are told that she visits san francisco and Denver, we will fix the two cities to be constant. Thus, we will have 6 cities, but permutate 4 of them. Which means, we have;

Ns = P(6, 4) = 6!/(6 - 4)! = 360

Thus;

P(visits Denver before San Francisco) = Ns/number of different itineraries that are possible = 360/720 = 0.5

55? i think maybe

step-by-step explanation:

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step-by-step explanation: