When trying to solve this problem I considered the principle of inclusion exclusion where the set U is the number of all possible arrangements, the set A is the number of arrangements that have "aa", the set B is the number of arrangements with "bb", and the set C is the number of arrangements with "cc." Then
$$N(U) = P(9;3,3,3),$$ $$N(A)=N(B)=N(C)=P(8;3,3,1),$$ $$N(A \cap B) = N(A \cap C) = N(B \cap C) = P(7;3,1,1),$$ $$ N(A \cap B \cap C) = 6!$$
Using $N(U)-N(A \cup B \cup C)$ and the inclusion exclusion principle I obtained the answer:
$$ \frac{9!}{3!3!3!} - 3 \frac{8!}{3!3!} + 3 \frac{7!}{3!} - 6!$$
as the number of arrangements without two adjacent letters the same.
However, does the fact that their is three of each letter impact my approach? Any advice would be appreciated!
