Australian Mathematics Competition 2022 Junior Level Question 25

Question

In a 2x3 grid, the numbers 1 to 6 are placed so that when joined in ascending order, they make a trail. The trail moves from one square to an adjacent square but does not move diagonally. In how many ways can the numbers 1 to 6 be placed in the grid to give such a trail.

So you could start from any point and label that as 1. Then there are 2 ways to reach another square. Then from one square you can make a technical path. Each possibility has 4 paths and there are 6 squares thus my answer was 4x6=24

But in fact the answer was 16 and I don't understand why this could be. If anyone can provide any guidance it would be greatly appreciated.

Thank you.

Answer 1

There are two squares in the middle column. If you start in one of those squares, there are only two possible acceptable paths: clockwise or counterclockwise. To see this, color the grid as a checkerboard and note that you must end either in an adjacent corner or in the middle column, and the latter choice is impossible because there’s no way to get to both corners without traversing the middle between them. That means the two adjacent corners must be $2$ and $6$ and the other middle column square must be $4$. That accounts for $2 \times 2 = 4$ possible paths.

There are $4$ corner squares. Again , using checkerboard coloring shows there are three possible endpoints for the path. Each is attainable in exactly $1$ way. (If the path doesn’t end at the adjacent corner, then since there are only two squares adjacent to it and one of them has been used, the adjacent corner must be $2$.) Thus, starting in a corner square accounts for $4 \times 3=12$ additional paths.

Therefore, there are a total of $4+12=16$ possible paths.

It is not correct that each square has $4$ possible paths. In fact, no starting point yields $4$ possible paths.

Answer 2

Have $(x,y)$ be the $1$-based position of the $x$'th row and $y$'th column. Consider the different locations where the number $1$ may be placed. If it's in the upper left corner, i.e., $(1,1)$, then there are $2$ possible positions for $2$. First, if it's $(1,2)$, then the position for $3$ can't be $(2,2)$ because then $4$ would be on either the left or right side, with the other side being blocked off. Thus, the only possibility is that $3$ be at $(1,3)$, with $4$, $5$ and $6$ then being at $(2,3)$, $(2,2)$ and $(2,1)$, respectively.

Second, if $2$ is at $(2,1)$ instead, then $3$ must be at $(2,2)$. Next, $4$, $5$ and $6$ are at $(1,2)$, $(1,3)$ and $(2,3)$, or at $(2,3)$, $(1,3)$ and $(1,2)$, respectively. This gives a total of $3$ possibilities for $1$ starting at $(1,1)$.

Next, if $1$ is at $(1,2)$, then $2$ can't be at $(2,2)$ as one side will later be blocked. Instead, the remaining values can go either clockwise around it (i.e., at $(1,3)$, $(2,3)$, $(2,2)$, $(2,1)$ and $(1,1)$), or counter-clockwise (i.e., at $(1,1)$, $(2,1)$, $(2,2)$, $(2,3)$ and $(1,3)$). This gives $2$ more possibilities.

By symmetry, having $1$ at $(1,3)$ gives the same # as from $(1,1)$, i.e., $3$. This gives a total of $3 + 2 + 3 = 8$ possibilities with $1$ being in the top row. Having $1$ in the second row instead will give the same number of possibilities. Thus, the total number of ways to place the numbers is $2(8) = 16$.