(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:
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)
This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook.