82] ON THE TRIADIC ARRANGEMENTS OE SEVEN AND FIFTEEN THINGS. 483 
triads having a letter in common, there shall be triads such as £/38, £ye, and rj/3e, 
7]jS) may easily be found; the system to be presently given of the triads of fifteen 
things would answer the purpose. And so would many other systems. 
Dropping the consideration of the order of the letters which form a triad, I pass 
to the case of a system • of fifteen letters, a, b, c, d, e, f g, h, i, j, k, l, m, n, o. 
It is possible in this case, not only to form systems of thirty-five triads containing 
every possible duad, but this can be done in such manner that the system of thirty- 
five triads can be arranged in seven systems of five triads, each of these systems 
containing the fifteen letters 1 2 . My solution is obtained by a process of derivation 
from the arrangements ab. ,cf. dg. eh and ab.cd.ef.gh as follows; viz. the triads are 
iab jac kaf lad mag nae oali 
icf jfb kbc Ice mch ncd ocg 
idg jde kdh Igb mbd ngf ofd 
ieh jlig kge Ihf mfe nhb obe 
and a system formed with i, j, k, l, m, n o, which are then arranged in the form 
klo 
ino 
jmo 
ilm 
jin 
ijk 
kmn 
iab 
jac 
lad 
nae 
kaf 
mag 
oah 
ncd 
mdb 
kbc 
ocg 
mch 
Ice 
icf 
mef 
keg 
ieh 
jf h 
obe 
ofd 
jde 
J9 h 
Ihf 
n fd 
khd 
idg 
nhb 
Ibg 
an arrangement, which, it may be remarked, contains eight different systems (such as 
have been considered in the former part of this paper) of seven letters such as i, j, 
k, l, m, n, o; and seven of other seven letters, such as i, j, k, a, b, c, f(f). The 
theory of the arrangement seems to be worth further investigation. 
Assuming that the four hundred and fifty-five triads of fifteen things can be 
arranged in thirteen systems of thirty-five triads, each system of thirty-five triads 
containing every possible duad, it seems natural to inquire whether the thirteen 
systems can be obtained from any one of them by cyclical permutations of thirteen 
letters. This is, I think, impossible. For let the cyclical permutation be of the letters 
a, b, c, d, e, f g, h, i, j, k, l, m. Consider separately the triads which contain the 
letter n and the letter o; neither of these systems of triads contains the letter, 
whatever it is, which forms a triad with n and o. Hence, omitting the letters n, o, 
we have two different sets, each of them of six duads, and composed of the same 
twelve letters. And each of these systems of duads ought, by the cyclical permutation 
1 The problem was proposed by Mr Kirkman, and has, to my knowledge, excited some attention in the 
form “To make a school of fifteen young ladies walk together in threes every day for a week so that each 
two may walk together.” It will be seen from the text that I am uncertain as to the existence of a 
solution to the further problem suggested by Mr Sylvester, “ to make the school walk every week in the 
quarter so that each three may walk together.” 
2 [I have somewhat altered this sentence so as to express more clearly what appeared to be the meaning of it.] 
61—2