A Boring Day for Bridget Version 2.0

There are $2^j \times (12-j)!$ ways of ordering the shoes so that $j$ specific pairs are next to each other --- simply permute the $j$ pairs (each pair as a unit) and the $12-2j$ other shoes, and then multiply by $2^j$ since each of the particular pairs could be in one of two orders. By the Inclusion-Exclusion Principle, there are $\sum_{j=1}^6 (-1)^{j-1} \binom{6}{j} \times 2^j (12-j)!$ shoe orders with at least one pair together, and hence there are $\begin{aligned} 12! - \sum_{j=1}^6 (-1)^{j-1} \binom{6}{j} \times 2^j (12-j)! & = \; \sum_{j=0}^6 (-1)^{j} \binom{6}{j} \times 2^j (12-j)! \\ & =\; \boxed{168422400} \end{aligned}$ shoe orders where no pair is together.