probability pls try

nishant, i am getting something which is very different from the answers posted here...
my working:
we can get (6ⁿ) lists of elementary events, each of length (n). out of all these elementary events, we need to select that list which has only three numbers out of {1,2,3,4,5,6} ...these 3 numbers can be selected in ⁶C₃ ways !
...now by using three selected numbers, we can get 3ⁿ lists of elementary events , each of length (n). but these 3ⁿ lists may contain those lists which contain exacltly two numbers and exactly one number
the no. of lists of length (n) which contain exactly 2 nos. out of 3 chosen nos. = ³C₂(2ⁿ -2)
the no. of lists of length (n) which contain exactly 1 no. out of 3 chosen nos. =³C₁(1ⁿ) =3
thus the total number of lists of length (n) which contain three distinct nos. out of the six given numbers is ⁶C₃{3ⁿ- ³C₂(2ⁿ -2) -3 }
= ⁶C₃{3ⁿ - 3.2ⁿ +3}

hence required probability is [⁶C₃{3ⁿ - 3.2ⁿ +3}] / 6ⁿ