Henry Ernest Dudeney/Modern Puzzles/102 - The Nine Barrels/Solution
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?
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
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $102$. -- The Nine Barrels
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $138$. The Nine Barrels