Mr. W. Spottiswoode on a Problem in Combinatorial Analysis. 353 



must be also odd, and so on. In other words, in order that the 

 present method may be applicable, the operation of subtracting 

 unity from the given number and dividing by two carried on 

 successively, must never lead to an even number. From this it 

 will be seen that n must belong to the series 



1 



2.1 + 1 



2(2.1 + 1) + 1 



2(2(2. 1 + 1) + 1)4-1. 



And, since the number n must also be divisible by 3, it can 

 belong only to the even places (*. e. 2nd, 4th, . . ) in this series. 

 The numbers will be found to be 



3, 15, 63, 255, 1023, 4095, . . ; 



the 2j»th term being 



2 2p - l +2 2p - 2 + .. + 2 + 1 = 2^-1, 



which, as is well known, is always divisible by 2+1 = 3, thus 

 satisfying that condition of the problem. 



In order to show that the number of ternary combinations 

 formed in the manner above indicated is exactly sufficient for a 

 solution of the problem, it may be remarked that the number of 



combinations in each triangular arrangement is — ^ — ~, if m 



be the number of terms in the top row ; and consequently, the 

 numbers of ternary combinations will be 



(2 2p - 1 -l)2 2p - 2 = 2 ip - 3 -2 2p - 2 in the 1st triangle, 



(2 2p - 2 -l)2 2p - 3 = 2 ip - 5 -2 2p - 3 in the 2nd triangle, 



(2 2 — 1)2 =2 3 -2 in the (2p — l)th triangle, 



(2-1)1 =2-1 in the (2p - 2)th triangle, 



the sum of which is 



Q4/>-2_ 1 1 



2 . 1 _ 2 2p ~ ' + 1 = i (2 ip ~ l -2 2p - 2 2p ~ ' + 1) 



= l(2 2p -\)(2 2p - l -l), 



1 

 i. e. the number of combinations is equal to the product of ^ of 



o 



the given number of ladies by the number of days on which the 



walks are to be taken, as it should be. 



The solution of the general case here contemplated would 



Phil. Mag. S. 4. Vol. 3. No. 19. May 1852. 2 A 



