Is my reasoning for this probability question correct?: Four fair dice are rolled. What is the probability of getting two different pairs? Four fair dice are rolled. What is the probability of getting two different pairs?

I basically used the multiplication rule here: the first die can have $6$ possible outcomes, while the second one can have only $5$. The last two can only have one outcome each because they have to match the numbers on the first and second dice. Hence the numerator becomes $6 \cdot 5 \cdot 1 \cdot 1$. Then I divided by all possible outcomes, which is $6^4$. So the probability $= 30/6^4 = .023$.
I think that another way to solve the problem is to assume that the outcome of the four dice can be expressed as two pairs, one of them with 6 different outcomes (from 1 to 6) and the other one with just 5 different outcomes (because you can't repeat the outcome you got for the first pair). So, we get 30 results. Then, due to the fact that we will have outcomes that can be expressed as (XX)(YY) or (YY)(XX) which have the same results, we divide the total outcomes by 2, such that we have only outcomes composed by "unique pairs" (e.g. if you have an outcome (XX)(YY), you are not going to consider, by the moment, as a different outcome (YY)(XX)). In every outcome, you have 4 elements, but there are only two different pairs, so you will have $\frac{4!}{2!2!}=6$ different orderings. In the end, you will have $15\times6=90$ different outcomes that meet the rule, so the probability will be $\frac{90}{6^4}$ or 0.069.
Hint:

The event of getting two different pairs have three 'types'.


All four different
Three same, one different
Two same, other two different