ASSUMPTION: All the dice are fair.
$\frac{12!}{6^{6}\cdot 12^{6}}$ should be the right answer.
Event: Rolling $12$ dice.
Sample space: Every element of the sample space would be an ordered set of twelve numbers denoting outcomes of each of the twelve dice.We can think of an element of our sample space as a twelve character string where for each character we just have six different choices.
Since we have assumed that all the dice are fair, so every element of the sample space would have the same probability of happening as any other element in the sample space.
The number of elements in the sample space: Since each of the twelve dice can outcome six numbers, we can have $6^{12}$ different choices.So the cardinality of sample space would be $6^{12}$.
Now, all the elements of the sample space are equiprobable & we have $6^{12}$ elements in the sample sapce, and since the probability of all the elements in the sample space should add up to $1$ so the probability of happening of any single element from the sample space would be $\frac{1}{6^{12}}$.
Now, let A be a subset of sample space such that:
A : All the elements of sample space in which every possible outcome of a single die throw has exactly appeared twice, that is all the elements of sample space in which every number from the set $\left \{ 1, 2, 3, 4, 5, 6 \right \}$ has exactly appeared twice.
The number of elements in A would be same as the number of all twelve digit permutations in which every element from the set $\left \{ 1, 2, 3, 4, 5, 6 \right \}$ exactly appears twice.
Number of such permutations = number of elements in A $= \frac{12!}{2!\cdot 2!\cdot 2!\cdot 2!\cdot 2!\cdot 2!} = \frac{12!}{\left ( 2! \right )^{6}}$
Thus, the required probability = Probability of happening of A = (Probability of happening of a single element of A $\times$ Total number of elements in A) $= \frac{1}{6^{12}} \times \frac{12!}{\left ( 2! \right )^{6}} = \frac{12!}{6^{6}\cdot 12^{6}}$.
