Question

A businessman in New York is preparing an itinerary for a visit to seven major cities. The distance traveled, and hence the cost of the trip, will depend on the order in which he plans his route. How many different itineraries (and trip costs) are possible?

Answer 1

Answer:

The answer is 5,040.

There are 5,040 different possible itineraries.

Explanation:

The number of different possible itineraries equals the number of the selection of 7 cities from a total of 7 cities where order is important.

We solve thus:

$N = P^{7}_{7} = \frac{7!}{(7 - 7)!}$

   $= \frac{7!}{0!}$

   $= 7!$

   $= 7 X 6 X 5 X 4 X 3 X 2 X 1$

   $= 5,040$