I understand that the well known way to solve this is
- Choose the rank for the two pairs from 6 options - $6C2$
- Choose the dice for the first pair - $5C2$
- Choose the dice for the second pair - $3C2$
- the 5th roll has to be filled among the remaining $6-2=4$ options
So numerator is $6C2 \cdot 5C2 \cdot 3C2 \cdot 4 = 1800$
The sample space consists has $6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 = 6^5$ options
So probability of two pairs is $1800/6^5 = 0.2315$
However I fail to understand, in the denominator sample space, we do not consider the ordering of the dice, if we were to consider the ordering of the dice, the denominator should be $6^5 \times 5!$, then the numerator makes sense to me , as we consider the ordering of the dice via $5C2$ and $3C2$.
Where is my understanding going incorrect?