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

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Modern Puzzles by Henry Ernest Dudeney: $102$

The Nine Barrels
In how many different ways may these nine barrels be arranged in three tiers of three
so that no barrel shall have a smaller number than its own below it or to the right of it?
The first correct arrangement that will occur to you is $1 \ 2 \ 3$ at the top then $4 \ 5 \ 6$ in the second row, and $7 \ 8 \ 9$ at the bottom,
and my sketch gives a second arrangement.
How many are there altogether?
Dudeney-Modern-Puzzles-102-question.png


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.

$\blacksquare$


Sources

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