This is an extension of the well-known problem of the "Fifteen Schoolgirls," by Kirkman. The original conditions were simply that fifteen girls walked out on seven days in triplets without any girl ever walking twice in a triplet with another girl. Attempts at a general solution of this puzzle had exercised the ingenuity of mathematicians since 1850, when the question was first propounded, until recently. In 1908 and the two following years I indicated (see _Educational Times Reprints_, Vols. XIV., XV., and XVII.) that all our trouble had arisen from a failure to discover that 15 is a special case (too small to enter into the general law for all higher numbers of girls of the form 6n+3), and showed what that general law is and how the groups should be posed for any number of girls. I gave actual arrangements for numbers that had previously baffled all attempts to manipulate, and the problem may now be considered generally solved. Readers will find an excellent full account of the puzzle in W.W. Rouse Ball's _Mathematical Recreations_, 5th edition. 270.--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 13 x 14 x 15 x 16 ----------------- = 1,820 1 x 2 x 3 x 4 ways for A. In every one of these cases B may occupy any 4 out of the remaining 12 places, making 9 x 10 x 11 x 12 ----------------- = 495 1 x 2 x 3 x 4 ways. Thus 1,820 x 495 = 900,900 different placings are open to A and B. But for every one of these cases C may occupy 5 x 6 x 7 x 8 ------------- = 70 1 x 2 x 3 x 4 different places; so that 900,900 x 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. The following will be found to comply with the conditions of grouping:-- ALE MET MOP BLM BAG CAP YOU CLT IRE OIL LUG LNR NAY BIT BUN BPR AIM BEY RUM GMY OAR GIN PLY CGR PEG ICY TRY CMN CUE COB TAU PNT ONE GOT PIU