If two faces of a dice have a common edge, the two faces are called adjacent faces. In how many ways can we construct a dice with six faces such that any two consecutive numbers lie on two adjacent faces? 

$\textbf{(A)}\ 10  \qquad\textbf{(B)}\ 14  \qquad\textbf{(C)}\ 18  \qquad\textbf{(D)}\ 56  \qquad\textbf{(E)}\ \text{None}$
1. Let's start by placing the numbers \(1\) and \(2\) on the dice. We can visualize the dice in an unfolded (net) form as follows:

\[
\begin{array}{ccc}
  & 1 & \\
  & 2 & \\
A & B & C \\
  & D & \\
\end{array}
\]

2. We need to ensure that any two consecutive numbers lie on adjacent faces. Let's consider the placement of the number \(3\).

3. If \(A = 3\), then \(B = 4\) (since \(3\) and \(4\) must be adjacent). The remaining numbers \(5\) and \(6\) can be placed on \(C\) and \(D\) in two ways:
   - \(C = 5\) and \(D = 6\)
   - \(C = 6\) and \(D = 5\)

   Thus, there are \(2\) ways if \(A = 3\) and \(B = 4\).

4. If \(A = 3\) and \(D = 4\), then \(B\) and \(C\) can be either \(5\) or \(6\):
   - \(B = 5\) and \(C = 6\)
   - \(B = 6\) and \(C = 5\)

   Thus, there are \(2\) ways if \(A = 3\) and \(D = 4\).

5. Therefore, if \(A = 3\), there are \(2 + 2 = 4\) ways.

6. Similarly, if \(C = 3\), there are \(4\) ways (by symmetry).

7. If \(B = 3\), then \(D\) cannot be \(4\) because \(A\) and \(C\) are not adjacent. So, if \(A = 4\), \(D\) must be \(5\) and \(C\) must be \(6\). Alternatively, if \(C = 4\), then \(D = 5\) and \(A = 6\). Thus, there are \(2\) ways if \(B = 3\).

8. Summing up all the possibilities, we get:
   \[
   4 \text{ (if } A = 3) + 4 \text{ (if } C = 3) + 2 \text{ (if } B = 3) = 10
   \]

Conclusion:
\[
\boxed{10}
\]