Selection tests for the Balkan Mathematical Olympiad 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 $2\left(2^2+3^2\right)+4^2=42$.