(related to Problem: The Two Pawns)

Call one pawn $A$ and the other $B.$ Now, owing to that optional first move, either pawn may make either $5$ or $6$ moves in reaching the eighth square. There are, therefore, four cases to be considered:

  1. $A$ $6$ moves and $B$ $6$ moves;
  2. $A$ $6$ moves and $B$ $5$ moves;
  3. $A$ $5$ moves and $B$ $6$ moves;
  4. $A$ $5$ moves and $B$ $5$ moves.

In case (1) there are $12$ moves, and we may select any $6$ of these for $A.$ Therefore $7 \times 8 \times 9 \times 10 \times 11 \times 12$ divided by $1 \times 2 \times 3 \times 4 \times 5 \times 6$ gives us the number of variations for this case — that is, $924.$

Similarly for case (2), $6$ selections out of $11$ will be $462$:

In case (3), $5$ selections out of $11$ will also be $462.$

And in case (4), $5$ selections out of $10$ will be $252.$

Add these four numbers together and we get $2,100,$ which is the correct number of different ways in which the pawns may advance under the conditions. (See The Glass Balls)

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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