A salesperson must travel to four cities to promote a new marketing campaign. How many different trips are possible if any route between cities is possible?
How many different trips are possible?
A. 4
B. 4 !
C. $4^{4}$
D. 8

Step 1 ：A salesperson must travel to four cities to promote a new marketing campaign. How many different trips are possible if any route between cities is possible?

Step 2 ：The salesperson can start from any of the 4 cities. After choosing the first city, he has 3 options for the next city, then 2 options for the city after that, and finally only 1 option for the last city.

Step 3 ：This is a permutation problem, and the number of permutations of n distinct objects taken r at a time is given by nPr = n! / (n-r)!. Here, n = r = 4, so the number of different trips is 4! = 4*3*2*1.

Step 4 ：Substituting the values, we get n = 4 and r = 4.

Step 5 ：Calculating the number of trips, we get number_of_trips = \(4! = 4 \times 3 \times 2 \times 1 = 24\).

Step 6 ：Final Answer: The number of different trips possible is \(\boxed{24}\). So, the correct option is B. 4 !.