(related to Problem: The Glass Balls)

There are, in all, sixteen balls to be broken, or sixteen places in the order of breaking. Call the four strings $A, B, C,$ and $D$ — order is here of no importance. The breaking of the balls on $A$ may occupy any $4$ out of these $16$ places — that is, the combinations of $16$ things, taken $4$ together, will be

$$\frac{13 \times 14 \times 15 \times 16}{1 \times 2 \times 3 \times 4} = 1,820$$

ways for $A.$ In every one of these cases $B$ may occupy any $4$ out of the remaining $12$ places, making $$\frac{9 \times 10 \times 11 \times 12}{1 \times 2 \times 3 \times 4} = 495$$ ways. Thus $1,820 \times 495 = 900,900$ different placings are open to $A$ and $B.$ But for every one of these cases $C$ may occupy

$$\frac{5 \times 6 \times 7 \times 8}{1 \times 2\times 3 \times 4} = 70$$

different places; so that $900,900 \times 70 = 63,063,000$ different placings are open to $A, B,$ and $C.$ In every one of these cases, $D$ has no choice but to take the four places that remain. Therefore the correct answer is that the balls may be broken in $63,063,000$ different ways under the conditions. Readers should compare this problem with The Two Pawns, which they will then know how to solve for cases where there are three, four, or more pawns on the board.

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Project Gutenberg

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

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