1. A square table is divided into $16 \times 16$ cells. A knight moves on it in two directions: to the right or down, alternating jumps of two and three cells (that is, no two consecutive jumps are of the same length). It starts with a jump of length two from the top-left cell. In how many different ways can the knight reach the bottom-right cell? (A path is understood as a sequence of cells the knight jumps to.)
SOLUTION. In the course of its journey, the grasshopper must move a total of 15 squares to the right and 15 squares down. In total, it will move 30 squares, so it will repeat the pair of jumps of length $2+3=5$ six times. More precisely, its individual jumps will have lengths in sequence

$$
2,3,2,3,2,3,2,3,2,3,2,3 \text {, }
$$

so it will make six 2-jumps (jumps of length 2) and six 3-jumps (jumps of length 3). If we assign ordinal numbers to the individual 2-jumps and 3-jumps according to their position in (1), the grasshopper's path will be uniquely determined by selecting the ordinal numbers of the jumps heading to the right (the remaining ones will head down). We only need to ensure that the sum of the lengths of the selected jumps (i.e., the jumps to the right) equals 15. This can be achieved with the allowed lengths (without considering the order of the jumps) in the following ways:

$$
\begin{aligned}
& 15=3+3+3+3+3, \\
& 15=3+3+3+2+2+2, \\
& 15=3+2+2+2+2+2+2 .
\end{aligned}
$$

In the first case, five of the six 3-jumps will be to the right (and all 2-jumps will be down), so the path will be determined only by the ordinal number of the (single) 3-jump that will go down. Therefore, there are 6 paths of this type.

In the second case, the path will be determined by the ordinal numbers of the three 3-jumps to the right and the ordinal numbers of the three 2-jumps to the right. The selections of both triplets are independent (i.e., they can be combined in any way), and in each case, we are selecting three elements from six, which can be done in 20 ways. ${ }^{1}$ Therefore, there are $20 \cdot 20=400$ paths of this type.

In the third case, the grasshopper's path is determined only by the ordinal number of the single 3-jump that will go to the right, so there are (as in the first case) again 6 paths of this type.

Answer: The total number of the grasshopper's paths is $6+400+6=412$.