jmc

Let us partition the set $\{1,2,\cdots ,12\}$ into $n$ numbers in $A$ and $12-n$ numbers in $B$, Since $n$ must be in $B$ and $12-n$ must be in $A$ ($n\ne 6$, we cannot partition into two sets of 6 because $6$ needs to end up somewhere, $n\ne 0$ or $12$ either). We have $\left(\genfrac{}{}{0ex}{}{10}{n-1}\right)$ ways of picking the numbers to be in $A$. So the answer is $\left({\sum }_{n=1}^{11}\left(\genfrac{}{}{0ex}{}{10}{n-1}\right)\right)-\left(\genfrac{}{}{0ex}{}{10}{5}\right)=2^{10}-252=\boxed{772}$.