GMAT Prep Roller Coaster

3!/3^3

=6/27 =2/9 C

Hey Samir dont mind but can you please explain how you arrived at the answer because I am not that good at probability questions..

let the cars be represented by A,B,C

so total possibilities = 3*3*3 (each time he can select one of the 3 cars)

& for him to select each of the cars A,B,C, he can do this using all permutations of ABC i.e 3!

so reqd P =3!/3^3 =2/9

Sorry Moneyman for not providing the expln, actually I'm trying to practise solving Q's on the scratch paper. 8)