Given the alphabet $\Sigma := \{ a, b, c \}$, find the number of all possible length-$4$ words over $\Sigma$.
In this case, the order does of course matter and it is also the case that same letter can appear multiple times in the word. So the total amount of possibilities is $3^4$.
What I do not understand is the following case. Let's say I have the word $abbc$ and then I switch the two $b$'s with each other and become $abbc$ but the first $abbc$ is the same as the second $abbc$, so in this specific case the order doesn't matter? But when forming words we say that the order does matter. Does anyone see where it went wrong in my explanation?