Given are a rectangular board of size $13 \times 2$ and arbitrarily many dominoes of sizes $2 \times 1$ and $3 \times 1$. The board is to be completely covered by such dominoes without gaps or overlaps, and no domino is allowed to protrude beyond the board. Furthermore, all dominoes must be aligned in the same direction, i.e., their long sides must be parallel to each other.

How many such coverings are possible?

(Walther Janous)

Answer. There are 257 possible coverings.
For the sake of terminological simplicity, we assume that the longer side of the playing field lies horizontally.

If the longer sides of the dominoes are placed parallel to the short side of the playing field next to each other, only those of size $2 \times 1$ can be used, and all tiles must be placed vertically. For this covering, there is exactly one possibility.

If the longer sides of the dominoes are placed parallel to the long side of the playing field next to each other, all tiles must be placed horizontally. Let $x$ and $y$ be the respective number of dominoes of size $2 \times 1$ and $3 \times 1$ used to cover a row of the playing field. Then $2 x + 3 y = 13$ must hold. One immediately finds the only non-negative solutions to this Diophantine equation to be $x=2$ and $y=3$ or $x=5$ and $y=1$. Thus, for each row of the playing field, either $2+3=5$ or $5+1=6$ dominoes can be used. Different coverings of the row arise by the different orders in which the $2 \times 1$ and $3 \times 1$ dominoes lie on the playing field. The number of these orders corresponds in the first case to the number of ways to choose 2 positions for the $2 \times 1$ dominoes out of 5 possible positions, thus $\left(\begin{array}{l}5 \\ 2\end{array}\right)=10$. Analogously, there are $\left(\begin{array}{l}6 \\ 5\end{array}\right)=6$ ways to distribute the five $2 \times 1$ dominoes over the 6 positions in the second case. In total, there are thus 16 possibilities per row. Since the coverings in the two rows can be performed independently of each other, we get $16^{2}=256$ possible coverings for both rows.

In total, there are thus $1+256=257$ coverings possible.

(Reinhard Razen)