, 2013

Let us count the number of five-point sequences according to the position of point $P_{3}$.

Because the horizontal line of equation $y=3$ is a symmetry axis for this figure, the number of sequences with $x_{3}=i$ is equal to the number of sequences with $x_{3}=6-i$, for $i=1,2$.

Similarly, the vertical line of equation $x=3$ is a symmetry axis for this figure, so that the number $N_{i}$ of three-point sequences $\left(P_{1}, P_{2}, P_{3}\right)$ ending at $P_{3}$ with $x_{3}=i$ is equal to the number of three-point sequences $\left(P_{3}, P_{4}, P_{5}\right)$ starting at $P_{3}$, with $x_{3}=i$, for $i=1,2,3$. Therefore, the number of five-point sequences $\left(P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\right)$ with $x_{3}=i$ is $N_{i}^{2}$. Hence, the total number of five-point sequences $\left(P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\right)$ is
$$2\left(N_{1}^{2}+N_{2}^{2}\right)+N_{3}^{2} .$$
It is easy to see that $N_{1}=2$, since the only possibilites for $x_{4}, x_{5}$ are $x_{4}=2$ and $x_{5}=1$ or $3$.

It is also easy to see that $N_{2}=3$, since $x_{4}=1$ or $3$ and $x_{5}=2$ in the first case and $2$ or $4$ in the second case.

Finally, $N_{3}=4$, since $x_{4}=2$ or $4$, and $x_{5}=1$ or $3$ in the first case, and $3$ or $5$ in the second case.

Hence, the number of sequences is $2\left(2^{2}+3^{2}\right)+4^{2}=42$.