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.

[271.—FIFTEEN LETTER PUZZLE.—solution]

The following will be found to comply with the conditions of grouping:—

ALEMETMOPBLM
BAGCAPYOUCLT
IREOILLUGLNR
NAYBITBUNBPR
AIMBEYRUMGMY
OARGINPLYCGR
PEGICYTRYCMN
CUECOBTAUPNT
ONEGOTPIU