Solution:

The answer is $\mathbf{(E)}$. The required pairs can be constructed as follows: first we choose the common element between $A$ and $B$ (five possibilities), and for each element not belonging to the intersection we decide whether it lies in $A$, in $B$, or in neither of the two. This leads us to making a choice among 3 possibilities for each of the other 4 elements, so in total we have $3^{4}$ possibilities (once the intersection has been fixed). Since every pair with the required property is obtained in this way for exactly one choice of the intersection element and for exactly one choice of how to distribute the remaining elements, the number of pairs is $5 \cdot 3^{4} = 405$.