Let us solve a more general problem of determining the number $a_n$ of all coverings of a chessboard $3 \times 2n$ by pieces $2 \times 1$, for a given natural $n$. We will attack the problem by a recursive method, starting with $n = 1$.

The value $a_1 = 3$ (for the chessboard $3 \times 2$) is evident (see Fig. 2). To prove that $a_2 = 11$ by a direct drawing all possibilities is too laborious. Instead of this, we introduce new numbers $b_n$: Let each $b_n$ denote the number of all "incomplete" coverings of a chessboard $3 \times (2n - 1)$ by $3n - 2$ pieces $2 \times 1$, when a fixed corner field $1 \times 1$ (specified in advance, say the lower right one) remains uncovered. Thanks to the axial symmetry, the numbers $b_n$ remain to be the same if the fixed uncovered corner field will be the upper right one. Moreover, it is clear that $b_1 = 1$.

Fig. 2

Now we are going to prove that for each $n > 1$, the following equalities hold:
$$b_n = a_{n-1} + b_{n-1} \quad \text{and} \quad a_n = a_{n-1} + 2b_n. \quad (1)$$

Fig. 3

Similarly, the second equality in (1) follows from a partition of all coverings of a chessboard $3 \times 2n$ into three (disjoint) classes which are formed by coverings of types C, D and E respectively, see Fig. 4. It is evident that the numbers of elements in the three classes are $a_{n-1}$, $b_n$ and $b_n$, respectively.

Fig. 4

Now we are ready to compute the requested number $a_5$. Since $a_1 = 3$ and $b_1 = 1$, the proved equalities (1) successively yield
$$\begin{aligned} b_2 &= a_1 + b_1 = 4, \quad a_2 = a_1 + 2b_2 = 11, \quad b_3 = a_2 + b_2 = 15, \quad a_3 = a_2 + 2b_3 = 41, \\ b_4 &= a_3 + b_3 = 56, \quad a_4 = a_3 + 2b_4 = 153, \quad b_5 = a_4 + b_4 = 209, \quad a_5 = a_4 + 2b_5 = 571. \end{aligned}$$

*Answer.* The number of coverings of the chessboard $3 \times 10$ equals $571$.