November 1, 1886.] 



♦ KNO^A^LEDGE ♦ 



13 



running on the numerals, fresh columns being added at the 

 right, then bj- setting a strip bearing the lettei^s a, b, c, ic, 

 as in the lowest row of the figure above, at whatever dis- 

 tance from the top the number of schoolgirls or chess- 

 players may require, we have at once the solution for that 

 number. 



The number of letters in this method will be odd, and 

 the letter afterwards added making the total number even. 

 Obviously, if the total number is 2ii, there will be n pairs 

 each (lay, and (2;* — 1) days of difterent arrangements. 



With regard next to the tifteen schoolgirls. In the 

 numbers above referred to several methods are given for 

 dealing with problems of the kind, where the division is into 

 sets of three. But the 1.5 may be dealt ivith thus : — - 



^Ye i-equire all the 15 to go out, each day, in threes, for 

 seven days. There will thus be 35 sets of thi-ee, no two of 

 one set being found in any other. If we can find such 3-5 

 sets, we shall have solved the problem, for it will then be 

 easy to t;ike these sets five and five, each five containing all 

 the 15 girls. This being so, we might solve the problem by 

 a hammer-and-tongs method — writing down all the 455 

 combinations, and then selecting 35 fulfilling the condition 

 indicated, and arranging these as required. Thus our sets 

 of combinations and our work of sifting out would begin 

 like this— 



Combinatious. Nnmber. Xo. Taken. 



ABC, ABD, ABE, ABO 13 ABC 



ACD, ACE, ACQ 12 none 



ADE, ADO 11 ADE 



AMN, A3I0 2 none 



ANO 1 ASO 



91 



55 



And so on to the end. (I have given the d combinations 

 more fully than the others, to show how as the end draws 

 near the combinations available have to be sought further 



along the rows.) Proceeding in this way we get the follow- 

 ing 35 combinations : — 



ABC 

 ADE 

 AFG 

 AHI 

 AJK 

 ALM 

 AXO 



BDF 

 BEG 

 BHJ 

 BIK 

 BLy 

 BMO 



CDG 

 CEF 

 CHK 

 CIJ 

 CLO 

 CMX 



GHO 

 GIN 

 GJM 



GKL 



These thirty-five combinations can be obtained more con- 

 veniently, however, by a method which was suggested in 

 Knowledge, Vol. I. (No. 9). (I fear the volume is out of 

 print.) The follo^ving is a simplified explanation. The 

 plan depends on putting down fii-st all the available com- 

 binations teginning with a, viz., abc, ade, afg, ahi, .vjk, 

 ALM, AXO ; then taking the fourteen lettei-s without a, and 

 out of them making all the available combinations beginning 

 with b and c, six of each ; then those beginning with D, E, 

 F, and G, four of each. The best way of doing this is 

 illustrated in the following figure, where the dots stand for 

 the letters similarly situated in the a column : — 



ABC 



H I 



NO 



B9 



I 

 11 

 II 



"X 

 X 

 X 



This gives the same arrangement as before. (The arrange- 

 ment should be compared with the above figure, when any 

 difliculty as to the interpretation of the dots will be at once 

 removed.) 



If we now make sets of fifteen out of these combinations 



of three, we do not at once get the arrangement required ; 



but each trial requires only a few minutes, and there are 



not more than three or four cases to be dealt with. We 



presently obtain the following arrangement : — 



1st day. 2nd day. 3rd day. 4th day. .5th day. 6th day. 7th day. 



ABC ADE AFG AHI AJK ALM ANO 



DIM BLN BMO BEG BDF BHJ BIK 



EJO CHK CIJ CLO CMX CEF CDG 



FHX FIO DHL DJN GHO GIN EHM 



GKL GJM EKX FKM EIL DKO FJL 



In like manner other cases can be dealt with where sets of 

 three are in question. The numbei-s of Knowledge above 

 mentioned contain general rules and methods for such cases. 

 The problem relating to the sixteen whist-players, who 

 are to sit down at four tables on five successive evening.s, so 

 that no two shall play twice in the same set of four, was of 

 my own invention. When I set it I had not solved it, but 

 I knew on general principles that it must admit of being 

 solved. It has proved a trifle tougher than I expected, but 

 is not very difficult. I defer the solution for a month, 

 because I find that it depends on another problem relating 

 to sets of three, which may be thus presented : — 



Problem. — How may twelve schoolyirh go out tline, and 

 three on four days, and then four and four on a fifth day (or 

 twelve whist-players lAay single dummy four days and the 

 full game on a fifth day, each day retainiiiy their ])laces un- 

 changed throughout), in such a way that no two will twice 

 belong to the same set whether triplet or giiadruplet ? 



The solution of this problem, as well as that of the problem 

 of sixteen whist-play ei^s, will be given next month. 



