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
13 × 14 × 15 × 16
----------------- = 1,820
1 × 2 × 3 × 4
ways for A. In every one of these cases B may occupy any 4 out of the
remaining 12 places, making
9 × 10 × 11 × 12
----------------- = 495
1 × 2 × 3 × 4
ways. Thus 1,820 × 495 = 900,900 different placings are open to A and B.
But for every one of these cases C may occupy
5 × 6 × 7 × 8
------------- = 70
1 × 2 × 3 × 4
different places; so that 900,900 × 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 No. 345, "The Two
Pawns," which they will then know how to solve for cases where there are
three, four, or more pawns on the board.