jmc

Since there will be $9$ elements after removal, and their mean is $6$, we know their sum is $54$. We also know that the sum of the set pre-removal is $66$. Thus, the sum of the $2$ elements removed is $66-54=12$. There are only $\boxed{\text{ 5}}$ subsets of $2$ elements that sum to $12$: $\{1,11\},\{2,10\},\{3,9\},\{4,8\},\{5,7\}$.