Number of possible results with rolls

Question

I have tomorrow a Combinatorics exam and I am practising with an exercise which asks me to find how many possible results are there if we throw $6$ identical dices so that there must be at least one three. I think is really simple, however, I'm not still sure:
One begins using inclusion-exclusion principle, which would be the cardinal of the total set $A$ of possible results minus the cardinal of another set $B$, which is the set of results in which no three appears. For the first cardinal, it is the same as putting the results in order and a bar between two numbers when those two numbers and different. If a certain number does not appear, we put two bars together. Therefore, one has $5$ bars to put into $(6+5) = 11$ positions, and therefore we have $11\choose 5$.
For the second cardinal, it is the same but taking into account that now there are $5$ possible numbers, therefore a bar less and a position less to choose, which leads to $10 \choose 4$. The final result would be ${11 \choose 5} - {10 \choose 4}$. Is this right?

Answer 1

Cardinal of $A$ is $6^6$, because each of the six dices has $6$ numbers on it. The cardinal of $B$ is $5^6$, because each of the six dices has $5$ favorable situations.

When throwing dices the order is important. For example, if we throw 2 dices, the pair $(1,2)$ is different of $(2,1)$, because the first position is for the first dice and the second position is for the second dice.