Example 4.2.11 Use $1 \times 2$ dominoes to cover a $4 \times 4$ chessboard, how many covering schemes are there.
Solution 1: Consider the middle line $AB$ of a $4 \times 4$ chessboard, which divides the $4 \times 4$ into two parts, each being $4 \times 2$.

Each part has an even number of squares, and any domino covering of the $4 \times 4$ that intersects with $AB$ must consist of an even number of dominoes.
(1) No domino intersects with $AB$, the total number of coverings is $F_{4} \times F_{4}=25$.
(2) Two dominoes intersect with $AB$, there are $4+1+1+4=10$ ways.
(3) Four dominoes intersect with $AB$, the covering is unique, so the total number of coverings is $25+10+1=36$.
Solution 2: Consider the $2 \times 2$ square in the middle of the $4 \times 4$.

(1) The $2 \times 2$ square contains two whole dominoes (Fig. $4-31$).

For each case, there are two possible coverings on the edges, so there are $2 \times 2=4$ in total.
(2) The $2 \times 2$ square contains one whole domino (Fig. 4-32).

There are four cases where the $2 \times 2$ square contains one whole domino. For each case, the two empty squares in the $2 \times 2$ can be covered by a domino in two ways (horizontal or vertical), and the rest is uniquely determined, so there are $4 \times 2 \times 2=16$ covering schemes in total.
(3) The $2 \times 2$ square does not contain any whole domino (Fig. $4-33$), each square in the $2 \times 2$ can be covered by a domino in two ways, and then it is uniquely determined, so there are $2^{4}=16$ covering schemes in total.
There are $4+16+16=36$ covering schemes in total.