2. Specimen of Auriferous Quartz. By Patrick Dudgeon.

The specimen was found near Wanlockhead, Dumfriesshire, in 1872, by Andrew Gemmell, miner, who unfortunately broke up the piece, and disposed of the fragments to different parties in the locality. Mr Dudgeon was enabled to get possession of all the fragments, and has restored the mass to its original form. The specimen is a very interesting one, being the largest known piece of auriferous quartz found in Scotland; and Mr Dudgeon thought it would interest the fellows of the Society to exhibit it at the meeting before placing it in the Museum of Science and Art, which has now been done with the consent of the different owners.

The following Gentlemen were elected Fellows of the Society:

GEORGE BROADRICK, C.E., Claremont Cottage, Leith.

JOHN NAPIER, Engineer, Lancefield House, Glasgow.
JAMES KING, of Campsie, 12 Claremont Terrace, Glasgow.
GEORGE CUNNINGHAM, C.E., 2 Ainslie Place.

Monday, 19th March 1877.

SIR WILLIAM THOMSON, President, in the Chair.

The following Communications were read:

1. On a Problem of Arrangements. By Professor Cayley. It is a well-known problem to find for n letters the number of the arrangements in which no letter occupies its original place; and the solution of it is given by the following general theorem :viz., the number of the arrangements which satisfy any r conditions is

where 1 denotes the whole number of arrangements; (1) the number of them which fail in regard to the first condition; (2) the number which fail in regard to the second condition; (12) the number which fail in regard to the first condition, and also in

...

regard to the second condition; and so on: (1) means (1) + (2) +(r): (12) means (12) + (13) + (2r) . . . + (r−1, r); and so on, up to (12. ... r), which denotes the number failing in regard to each of the r conditions.

Thus, in the special problem, the first condition is that the letter in the first place shall not be a; the second condition is that the letter in the second place shall not be b; and so on; and taking r=n, we have the known result, No. =

...

n=2,3,4, 5, 6, 7, ..

No. 1, 2, 9, 44, 265, 1854

I proceed to consider the following problem, suggested to me by Professor Tait, in connexion with his theory of knots: to find the number of the arrangements of n letters a b c ..jk, when the letter in the first place is not a or b, the letter in the second place not b or c, the letter in the last place not k or a. Numbering the conditions 1, 2, 3 .. n, according to the places to which they relate, a single condition is called [1]; two conditions are called [2] or [1, 1], according as the numbers are consecutive or non-consecutive: three conditions are called [3], [2, 1] or [1, 1, 1], according as the numbers are all three consecutive, two consecutive and one not consecutive, or all non-consecutive; and so on the numbers which refer to the conditions being always written in their natural order, and it being understood that they follow each other cyclically, so that 1 is consecutive to n. Thus, n = 6, the set 126 of conditions is [3], as consisting of 3 consecutive conditions; and similarly 1346 is [2, 2].

Consider a single condition [1], say this is 1; the arrangements which fail in regard to this condition are those which contain in the first place a or b; whichever it be, the other n 1 letters may

be arranged in any order whatever; and there are thus 2 II(n − 1) failing arrangements.

Next for two conditions; these may be [2], say the conditions are 1 and 2, or else [1, 1] say they are 1 and 3. In the former case the arrangements which fail are those which contain in the first and second places ab, ac, or bc, and for each of these the other n-2 letters may be arranged in any order whatever; there are thus 3 II (n-2) failing arrangements. In the latter case the failing arrangements have in the first place a or b, and in the third place c or d,-viz., the letters in these two places are a. c, a.d, b.c, or b.d, and in each case the other n 2 letters may be arranged in any order whatever: the number of failing arrangements is thus 2.2.II (n-2). And so in general when the conditions are [a, B, y..], the number of failing arrangements is = (a + 1)(B+ 1)(y + 1) . . II(n − a − ẞ-y..). But for [n], that is for the entire system of the n conditions, the number of failing arrangements is (not as by the rule it should be n+1, but) = 2, -viz., the only arrangements which fail in regard to each of the n conditions are (as is at once seen), abc..jk, and be...jka.

=

=

=

Changing now the notation so that [1], [2], [1,1], &c., shall denote the number of the conditions [1], [2], [1,1], &c., respectively, it is easy to see the form of the general result: if, for greater clearness, we write n = 6, we have

+ Σ(12)

No. 720 ([1]=6)2} 120+ ([2] = 6)3 24

=

+ ([1,1] = 9)2.2ƒ

The formula for the next succeeding case, n = 7, gives

= 80.

=

No. 5040 - 10080+ 9240 - 5040 +1764 - 392 + 49 -2,579.

Those for the preceding cases, n = 3, 4, 5, respectively are

We have in general [1]=n, [2]=n, [1, 1] = {n(n − 3); and in the several columns of the formulæ the sums of the numbers thus represented are equal to the coefficients of (1+1)2, thus, n = 6 as above, the sums are 6, 15, 20, 15, 6, 1. As regards the calculation of the numbers in question, any symbol [a, ẞ, y] is a sum of symbols [a-a+B-B'+y-y'. .], where a'+ẞ'+y.. is any partition of n-(a+ẞ+y..); read, of the series of numbers 1, 2, 3 ... n, taken in cyclical order beginning with any number, retain a, omit a', retain ẞ, omit B', retain y, omit y',.. Thus in particular, n = 6, [1, 1] is a sum of symbols [1-3+1-1] and [1-2+1-2]; it is clear that any such symbol [a'-a'+B-B'. .] is n or a submultiple of n (in particular if n be prime, the symbol is always =n): and we thus in every case obtain the value of [a, ß, y . .] by taking for the negative numbers the several partitions of n− (a+B+y..) and for each symbol [a-a'+ẞ- ß' + y − y'. .], writing its value, n or a given submultiple of n, as just mentioned. There would, I think, be no use in pursuing the matter further, by seeking to obtain an analytical expression for the symbols [a, ẞ, y . .].

*

=

For the actual formation of the required arrangements, it is of course easy, when all the arrangements are written down, to strike out those which do not satisfy the prescribed conditions, and so obtain the system in question. Or introducing the notion of substitutions, and accordingly considering each arrangement as derived by a substitution from the primitive arrangement abcd... jk, we can write down the substitutions which give the system of arrangements in which no letter occupies its original place: viz., we must for this purpose partition the n letters into parts, no part less than 2, and then in each set taking one letter (say the first in alphabetical order) as fixed, permute in every possible way the

* In explanation of the notation of substitutions, observe that (abcde) means that a is to be changed into b, b into c, e into d, d into e, e into a; and similarly (ab)(cde) means that a is to be changed into b, b into a, c into d, d into e,

e into c.

other letters of the set; we thus obtain all the substitutions which move every letter. Thus n=5, we obtain the 44 substitutions for the letters abcde, viz., these are

(abcde), &c., 24 symbols obtained by permuting in every way. the four letters b, c, d, e.

(ab) (cde), &c., 20 symbols corresponding to the 10 partitions ab, cde, and for each of them 2 arrangements

such as cde, ced.

And then if we reject those symbols which contain in any () two consecutive letters, we have the substitutions which give the arrangements wherein the letter in the first place is not a or b, that in the second place not b or c, and so on. In particular n=5, rejecting the substitutions which contain in any ( ), ab, bc, cd, de, or ea, we have 13 substitutions, which may be thus arranged :(acbed), (ac)(bed), (acebd), (adbec), (aedbc),

(aedbc), (bd)(aec),

(acedb), (ce)(adb),

(aecbd), (ad)(bec),

(adceb), (be)(adc).

Here in the first column performing on the symbol (acbed) the substitution (abcde), we obtain (bdcae), = (aebdc), the second symbol; and so again and again operating with (abcde) we obtain the remaining symbols of the column; these are for this reason said to be of the same type. In like manner symbols of the second column are of the same type; but the symbols in the remaining three columns are each of them a type by itself; viz., operating with (abcde) upon (acebd) we obtain (bdace), = (acebd); and the like as regards (adbec) and (aedbc) respectively. The 13 substitutions are thus of 5 different types, or say the arrangements to which they belong, viz.,

cebad, ceabd, cdeab, deabc, eabcd,

edacb, edabc,

caebd, daebc,

edbac, debac,

daecb, deacb.

are of 5 different types. The question to determine for any value of n, the number of the different types, is, it would appear, a difficult one, and I do not at present enter upon it.