5. On an $8 \times 8$ board, several dominoes (i.e., $2 \times 1$ rectangles) can be placed without overlapping each other. Let $N$ be the number of ways to place 32 dominoes, and $F$ be the number of ways to place 16 tokens on this board (no more than one token per cell). Which is greater, $N$ or $F$? Ways that can be transformed into each other by rotation or reflection of the board are considered different.
Solution. We will prove that the number of arrangements of 16 chips is greater than the number of partitions into 32 dominoes. To do this, we will devise a rule that associates each partition into 32 dominoes with an arrangement of 16 chips, such that different partitions correspond to different arrangements. In other words, we need to encode the partition with an arrangement in such a way that it can be uniquely restored.

Let's describe one such correspondence. Choose the less popular direction of the dominoes (vertical or horizontal). If we place a chip in the upper/left cell of each domino of the chosen direction, then (knowing which direction is chosen) it is easy to restore the position of all dominoes of that direction. After that, the other dominoes are also uniquely restored.

There are several difficulties here.

1) How can we determine which direction is chosen based on the arrangement of chips? Notice that the bottom-right cell of the board is always free. Let a chip in this cell indicate that the chosen direction is vertical. (This means that if there are 16 dominoes in each direction, we should choose the horizontal direction as the "less popular" one to avoid using the 17th chip.)
2) The described algorithm may result in fewer than 16 chips, but we need exactly 16. We need to add the missing chips to free cells.

2a) Suppose there are 9 to 15 chips on the board. Let's assume, for definiteness, that the chosen direction of the dominoes is vertical. Then the cells in the bottom row (except the rightmost one) are definitely free. We will place the extra chips in them (up to seven). An example is shown in the figure (the blue chip indicates the chosen direction, the green one is extra).

![](https://cdn.mathpix.com/cropped/2024_05_06_6036b04067de4d7185a7g-3.jpg?height=342&width=1252&top_left_y=726&top_left_x=466)

2b) Suppose there are fewer than 9 chips on the board. Then we will divide the left upper $6 \times 6$ square of this board into 9 squares of $2 \times 2$. At least one of them will be free (and if there are fewer than 7 chips, at least three will be free). We will occupy some of the free squares (for example, starting with the topmost of the leftmost) with chips entirely - so that there are at least 9 chips on the board, after which we will place the remaining chips as in point (a). Note that other parts of the algorithm cannot lead to the filling of an entire $2 \times 2$ square in the upper left fragment of $6 \times 6$, so during decoding, such squares are uniquely identified as "chip storage." An example is shown in the figure below (purple and green chips are extra).

![](https://cdn.mathpix.com/cropped/2024_05_06_6036b04067de4d7185a7g-3.jpg?height=323&width=1198&top_left_y=1595&top_left_x=493)

3) There are ways to arrange 16 chips that cannot result from the operation of our algorithm. For example, these are arrangements where there are filled "chip storage" areas, as described in point 2b, but the upper left $2 \times 2$ square is empty.

Thus, we have described an encoding algorithm that allows for unique decoding, so $N \leqslant F$. Since there are arrangements of 16 chips that are not generated by this algorithm, $N < F$.

Remark. Another approach is also possible, as described in the solution to problem 5 for grades 8-9: to make the correspondence simpler and not make it one-to-one, but to prove that each partition corresponds to more arrangements than each arrangement to partitions.