ab, ba, ca, da, ea, ac, bc, cb, db, eb, Note, When nism, or is m by in, or m-1 by m-1 is = I m-i the Number of &c. taken n by n, that mxm-IX m -2X Px pix p 2 X 9× P-3x &c. xqxq-1xq-2xq-3 × c. xrxr-ix &c. continued to m or m I Places r-2x-3x &c. x &c. continued to p, q, r, &c. Places respectively which is a more regular and fimple Series than thofe that are, or may be, exhibited by the former Method. CHA P. II. Of the Combinations of Quantities. Definition. THE feveral Ways or different Cafes of taking any required Number of Things propos'd, without regarding their Order or Places, are called the Combinations of thofe Things. Scholium 1. From the Nature of Alternations and Combinations rightly confider'd and compar'd, it will appear that the Number of Alternations of m Things a b c de, &c. different from each other, taken n by n, is equal to the Number of Combinations of the faid m Things taken n by 22 multiplied by the Number of Alternations of 22 Things different from each other, taken n by n: But the Number of Alternations of the faid m Things taken n by 22 is, by the Corollary in the laft Chap..: mx:m2:X:12 ·3:× &c. continued to n Places: And the Number of Alternations of 22 Things different from each other taken 22 by n is, by the faid Corollary, n× : n − 1 : × : 1 2:X:12 3:x &c, conti * Dd 12 nued nued to 2 Places; therefore the Number of Combinations of the faid m Things taken n by n is = Let it be required to find the Number of Combinations of the five Things abc de different from each other taken three by three. 3 X 2 X I Number of Combinations required. abc, acd, ade, bed, bde, cde. It is eafier to find the Number of Alternations of m Things ap 1 c', &c. taken n by n by the first Method in the laft Chap. than to find the Number of their Combinations by the like Method: And, when " is any great Number, it would require an impracticable Calculation to find the Number of their Combinations or Alternations by that Method: Wherefore, In order to find the Number of Combinations and of Alternatiors of m Things a b c, &c. taken n by n, if n be — m orm-1, then the Number of Alternations required may be eafily had by the Remark in the latter Part of the last Chap. But if n be lefs than m-1, and that the Number of Combinations of fuch Things fo taken is not many; then write down all thefe Combinations diftinctly, and then the Number of Alternations of each of those Combinations taken 2 by 2 will be found by the faid Remark. In order to which it will be convenient, in fome Cafes, to obferve that the Number of Combinations of m Things taten n by n ( being less than m) is Number of Combina-1 tions of the faid m Things taken m-n by m-n; for the Things compofing each Combination of thefe taken from the faid m Things leave the Things compofing each Combination of the former. Thus Suppofe Suppofe it was required to find the Number of Combinations and of Alternations of the eight Things a3 3 c2 taken fix by fix. By the above Obfervation the Number of Combinations required is that of the Combinations of the 8 Things a3 b3 c2 taken 8 6 by 8 ·6, viz. 2 by 2: And the Combinations themselves of the faid 8 Things taken 2 by 2, are a2, ab, ac, b2, bc, and c2; confequently which Combinations, being in Number fix, are all the Combinations which the eight Things a3 b3 c2 taken fix by fix are capable of. Next it will be proper to take Notice that four, and no more, of thofe Combinations, viz. the 2, 3d, 4th, and 6th have the fame Indices; and therefore the Number of Alternations of one of them is equal to the Number of Alternations of any other of them. Now, (by the Remark in the latter Part of the laft Chap.) the Number of Alternations of the firft Combination, viz. 6,× 5 × 4×3X2XI a3 b3 taken 6 by 6 is = 3 X 2 XIX 3 X 2 X I 20. Alfo, the Number of Alternations of the fecond Combi nation a3 b2 c taken 6 by 6 is — 6 × 5 × 4 × 3 X 2 X I 3 X 2 XIX 2 XIX I 60 -Number of Alternations of a3 bc2, as alfo of a2 b3 c, and of a b3 c2, feverally, taken 6 by 6. And the Number of Alternations of the fix Things in the fifth Combination, viz. of a2 b2c2 taken 6 by 6 is = 6 × 5 × 4×3 X 2 X I Confequently the Number of Alternations of the eight Things aaa bbb cc taken fix by fix is 20+60×4+90=350. I believe it was from a due Confideration of the like Examples with this, that Col. Thornycroft deduc'd his Method or Theorem for finding the Numbers of Combinations and Alternations of any Quantities expos'd taken n by n; to which Method, it being inferted in Phil. Tranf. No 299. I refer fuch Mathematicians as have a Mind to know more of this Doctrine. PART *Dd2 PART XIX. Of Magick Squares. UEST. 1. The Numbers, 1, 2, 3, 4, 5, 6, 7, 8, and Qu 9 being given; tis required to place them in a Magick Order; viz. in a fquare Form, fo as counting each Rank up and down, as alfo from one Hand to the other, and Diagonal-wife; that thofe Ranks may be equal to each other. Suppofe it done and reprefented in its proper Form by the following Symbols thus plac'd; viz. The Sum of the propos'd Figures is 45; and 43 15 is cach Side or Rank (fuppofe) s. I Then la+e+i=s) 2b+e+b=sBy the Nature of the Quest. 3/c+e+8= 1-1-2+3 4a -|- b-|-c -|- 3e +i+h+g 3 S. 15a1b+c+ i +b-1-8=25, per Quest. 4 563e= 6÷÷÷3 7 e The Value of e being thus found to be 5, there remain eight Figures more, viz. 1, 2, 3, 4, 6, 7, 8 and 9 ; But which of thofe is equal to any corner Letter, as fuppose a, is to be further fought. Beginning therefore with the leaft Number 1; I say the corner Letter a, and confequently any corner Letter as c, i, g, cannot be equal to it: For, if a was = 1, then i fhould be9; and b+c=15—1—14 as alfo =d-\-g: 'But there remain no two Numbers (after 5, 1 and 9) whofe Sum is 14 but 6 and 8; therefore, if either of thefe Figures were b, the other would be c; and, then no Figure would remain for the Value of either d or g: Wherefore a is not =1; neither is i, nor, confequently, any corner Letter equal to 1 or 9. 2 May be a, as will appear farther on. 3 Cannot be equal to a; for, if it were, then i fhould be =7; and b+c=15—3—12, as alfo=d+g: But there remain no two Numbers (after 5, 3 and 7) whofe Sum is 12 but 8 and 4, which cannot answer to b and c, and d and g; wherefore a, or any other corner Letter is not 35 neither is i, nor confequently any other corner Letter = 7. From what hath been faid 'tis plain that (if the Question propos'd is capable of being folv'd) the corner Letters are all equal to even Numbers; wherefore, placing one of them, as 2 for a, i will be 8, and c must be either equal to 4 or 6; let it (viz. c) be = 4; theng-6; b = 9; d=7; f=3; and b=1: and fo the Square is compleated as required. 2 9 4 But if c were 6 (a being - 2 2); then g=4; b = 7; d=9; f=1; and 3: And then the Square will ftand, thus Quest. 2. The Numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 and 16 being given; 'tis required to place them in a Magick Order, viz. in a fquare Form fo as counting each Rank from one Hand to the other, as alfo up and down, and Diagonal-wife, that those Ranks may be equal to each other. Suppofe it done and reprefented in its proper Form by the following Symbols thus plac'd; viz. a b c d |