Henry Ernest Dudeney/Modern Puzzles/102 - The Nine Barrels/Solution

by : $102$

 * The Nine Barrels

Solution
There are $42$ different arrangements.

Proof
Barrels $1$ and $9$ are fixed.

Let us place the $2$ immediately below the $1$.

If the $3$ is then below the $2$, there are $5$ arrangements, as the $4$ is now also fixed:


 * $\begin{array}{ccc}

1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 4 & 6 \\ 2 & 6 & 7 \\ 3 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 4 & 5 \\ 2 & 6 & 8 \\ 3 & 7 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 4 & 6 \\ 2 & 5 & 7 \\ 3 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 4 & 6 \\ 2 & 5 & 7 \\ 3 & 8 & 9 \end{array}$

If the $3$ is put to the right of the $1$, there are $5$ arrangements with the $4$ below the $2$:


 * $\begin{array}{ccc}

1 & 3 & 5 \\ 2 & 6 & 7 \\ 4 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 5 \\ 2 & 6 & 8 \\ 4 & 7 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 6 \\ 2 & 5 & 7 \\ 4 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 6 \\ 2 & 5 & 8 \\ 4 & 7 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 7 \\ 2 & 5 & 8 \\ 4 & 6 & 9 \end{array}$

another $5$ arrangements with the $5$ below the $2$:


 * $\begin{array}{ccc}

1 & 3 & 4 \\ 2 & 6 & 7 \\ 5 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 4 \\ 2 & 6 & 8 \\ 5 & 7 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 6 \\ 2 & 4 & 7 \\ 5 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 6 \\ 2 & 4 & 6 \\ 5 & 7 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 7 \\ 2 & 4 & 8 \\ 5 & 6 & 9 \end{array}$

another $4$ arrangements with the $6$ below the $2$:


 * $\begin{array}{ccc}

1 & 3 & 4 \\ 2 & 5 & 7 \\ 6 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 4 \\ 2 & 5 & 8 \\ 6 & 7 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 7 \\ 6 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 8 \\ 6 & 7 & 9 \end{array}$

and another $2$ arrangements with the $7$ below the $2$:


 * $\begin{array}{ccc}

1 & 3 & 4 \\ 2 & 5 & 6 \\ 7 & 8 & 9 \end{array} \qquad \begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 6 \\ 7 & 8 & 9 \end{array}$

This is $21$ arrangements with the $2$ under the $1$.

By symmetry, there are another $21$ arrangements with the $2$ to the right of the $1$.

This accounts for all the arrangements.