Our strategy is to first place the $1 \times 5$ and the $1 \times 4$ tiles since their size restricts their location. We have three cases: - Case 1: first row. There are 4 ways to place the $1 \times 4$ tile. There is an empty cell next to the $1 \times 4$ tile, which can either be occupied by the $1 \times 1$ tile or the $1 \times 2$ tile (see diagram). In both cases, there are 2 ways to place the remaining two tiles, so this gives $4 \cdot 2 \cdot 2=16$ ways. - Case 2: middle row. There are 4 ways to place the $1 \times 4$ tile, and the $1 \times 1$ tile must go next to it. There are 2 ways to place the remaining two tiles, so this gives $4 \cdot 2=8$ ways. - Case 3: bottom row. This is the same as Case 1 up to rotation, so there are also 16 ways to place the tiles here. In total, we have $16+8+16=40$ ways to place the tiles.