gmatprep probability problem

[sam117]

A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the
3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride
the roller coaster 3 times, what is the probability that the passenger will ride in each of
the 3 cars?

A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1

This is how I solved this problem:

Probability that the person rides the first car but not the other two = (1/3)(2/3)(2/3) = 4/27
Similarly probability for riding second car and not other two = 4/27
Similarly probability for riding third car and not other two = 4/27

So total probability = 3(4)/27 = 12/27 = 4/9

But the OA is [spoiler]2/9[/spoiler]

Can anyone explain what did I do wrong? Thanks

[mathbyvemuri]

Let the three cars be denoted as A,B and C.
If the order is to be taken care:
For three trips, the possibilities are:
3-trips, only-1 car => AAA,BBB,CCC -> 3P1 = 3!/(3-1)! = 3 possibilities
3-trips, 2 cars => AAB,ABA,BAA,ABB,BAB,BBA,AAC,ACA,CAA,BBC,BCB,CBB.... -> 3*3P2 = 18 possibilities
3-trips, all 3 cars => ABC,ACB,BAC,BCA,CAB,CBA -> 3P3 = 3!/(3-3)! = 6 possibilities
All possibilities = 3+18+6 = 27
Probability of using all cars = 6/27 = 2/9

[sam117]

Let the three cars be denoted as A,B and C.
If the order is to be taken care:
For three trips, the possibilities are:
3-trips, only-1 car => AAA,BBB,CCC -> 3P1 = 3!/(3-1)! = 3 possibilities
3-trips, 2 cars => AAB,ABA,BAA,ABB,BAB,BBA,AAC,ACA,CAA,BBC,BCB,CBB.... -> 3*3P2 = 18 possibilities
3-trips, all 3 cars => ABC,ACB,BAC,BCA,CAB,CBA -> 3P3 = 3!/(3-3)! = 6 possibilities
All possibilities = 3+18+6 = 27
Probability of using all cars = 6/27 = 2/9

Ok i got what you did but i still do not understand why my method is wrong! Thanks for your help