3. (6 points) $A, B, C, D$ four people stay in four rooms numbered $1, 2, 3, 4$, with one person per room; so that $B$ does not stay in room 2, and $B, C$ two people require to stay in adjacent numbered rooms. The number of ways to arrange this is. $\qquad$
【Solution】Solution: According to the problem, $B$ does not stay in room 2, so $B$ can stay in rooms $1, 3, 4$. If $B$ stays in room 1, then $C$ can stay in room 2, and the remaining 2 people can be arranged in the other 2 rooms. In this case, there are $A_{2}^{2}=2$ situations.
If $B$ stays in room 3, then $C$ can stay in rooms 2 or 4, which gives 2 situations, and the remaining 2 people can be arranged in the other 2 rooms,

in this case, there are $2 \times A_{2}^{2}=4$ situations.
If $B$ stays in room 4, then $C$ can stay in room 3, and the remaining 2 people can be arranged in the other 2 rooms. In this case, there are $A_{2}^{2}=2$ situations,

There are a total of $2+4+2=8$ situations,
Therefore, the answer is 8.