taken n-1 by -1; that each Alternation by fuch Pofition of a produc'd, will confift of n things; and that all these Alternations in Number equal to the Number of Alternations of the faid m-1 things a b&c. taken 2-1 by -I n - are all the Alternations that can be made of the faid 9 m things abc' &c. taken n by n, which will have any a in the firft Place of each Alternation of 'em. For the fame Reafon's the Number of Alternations of the faid m things abc" &c. taken n by n, which will have ab2 b in the faid firft Place of each of 'em, is equal to the Number of Alternations of the m― I things ab91c" &c. taken n-1 by n—i. Alfo the number of Alternations of the faid m things a bac &c. taken n by n, which will have c in the faid firft Place of each of them, is equal to the Number of Alternations of the m-1 things a b c &c. taken n-1 by n &c. Wherefore the Number of Alternations of the m things a b c &c. taken n by n, is equal to the Number of Alternations of the m-1 things a -I r p- s b 2 c &c. The Number of Alternations of the m things ap b 9-3 The Number of Alternations of the m-1 things T-I &c. Taken n-1 by n−1. Q. E. D. SCHOLIUM. In order to find the Number of Alternations of things p b q c' &c. taken one by one, two by two, or three by three, &c. by the help of our Lemma: Let the Indices in the faid m things, which are each or Greater than 1, 2, 3, 4, &c. be supposed Tt number of the either Equal to, A, B, C, D, &c. respectively: Part XVIII. respectively: Then the Number of the Indices in the faid m things, which are each equal to 1, 2, 3, &c. is A-B, B-C, C-D, &c. refpectively. Then, 1. The Number of Alternations of the m things ab cr &c. taken one by one, is manifeftly = A. 2. The Number of Alternations of m things abc" &c. taken two by two, is by our Lemma A-Bx Number of Alternations of m― I things, wherein is equal to the Number of all the Indices, + Bx Number of Alternations of m-1 things, wherein A is equal to the Number of all the Indices Taken one by one. =A-B× A-1+ BxA, By Parag. 1. A× A-1+ B. 3. The Number of Alternations of m things aba c* &c. taken three by three, is, by our Lemma, A-Bx Number of Alternations of m-1 things, wherein A-1,and B are equal to the Number of the Indices which are each either equal to, or greater than 1, and 2 respectively, +B-Cx Number of Alternations of m-r things, wherein A, and B-1 are equal to the Number of the Indices which are each either equal to, or greater than 1, and 2 respectively, +Cx Number of Alternations of m1 things, wherein A, and B are equal to the Number of the Indices, which are each either equal to, or greater than 1, and 2 respectively. Taken two by two. A-BXA-1XA-2+B+B-CX AXA-1+B-1 +C× A×A-1 B+. By Parag. 2, AA-1×A-2 +34B-3B+C. 1 4. The Number of Alternations of m things a ba c", &ci taken four by four, is, by our Lemma, A-Bx Number, of of Alternations of m1 things, wherein A1, B, and C are equal to the Number of the Indices which are each either equal to, or greater than 1, 2, and 3 Refpectively, +B-CX Number of Alternations of m-1 things, where in A, B1, and C are equal to the Number of the Indices which are each either equal to, or greater than 1, 2 and 3 Refpectively, +C-DX Number of Alternations of m1 things, where in A, B, and C-1 are equal to the number of Indices, which are each either equal to, or greater than 1, 2, and 3 Refpectively, +Dx Number of Alternations of mr things, wherein A, B and C, are equal to the Number of the Indices which are cach either equal to, or greater than 1, 2, and 3 respectively; Taken three by three. A-BXA-1 × A−2 × 4~33 B × A − 1 − 3 B+C A× +B-CXAXA-1×A~2+3 AXB-1-3XB-1+C +C-DXAXA-1× A~ 2 + 3 AB+3 B+C + DX AXA-1× A-2+3 AB3 B+C; By Parag. 3 4×A-1×4-2 × A −3+6AB18 AB 9 B + AXA 2 IX CA 3 B2+4C4-4 C+ D. སྐ &c. Example. Let it be required to find the number of Alternations of 3b3c2 (that is, of the 8 things aaabbbcc) taken four by four. Here A is 3, B3, C2, Do; Wherefore, by the 4th Parag. of our Scholium, 3 × 2 × 1 × 0 (0) + 6 × 3a × 3 (162) -18×3×3(-162) + 9×3 (27) + 3×3 (27) + 4×2×3(24) -4x2(-8)+0-70; is the Number of Alternations of a3 3 c2 taken four by four. COROLLARY. 2 From what has been faid in this Scholium, it is plain, that the Number of Alternations of m things different from each other, as a b c d, &c. taken n by n is (Because in this Cafe, A is equal equal to'm, and B,C,D, &c. are each equal to 0)=4x-1× I X A-2 × A-3× A-4x &c. continued to n places=m x m−1 x m-2xm-3× m-4 x &c, continued to n Places. Examples. 1. Let it be required to find the Number of Alternations of four Things different from each other, as abcd taken four by four. 2. Let it be required to find the Number of Alternations of the five things abcde different from each other, taken two by two. ab, ba, ca, da, ea, ae, bc, cb, db, eb, ad, bd, cd, dc, ec, ac, be, ce, de, ed. Here mA is 5 and n = 2; Therefore, by our Coroll. 5x4 = 20 is the Number of Alternations of five things different from each other, taken two by two. Note, When nism, the Number of Alternations of mn things ab &c. taken n by n; that is, m by m is = 9 Ixm-2 x m−3 × m − 4 × M-5 x m → 6xm- 7x p x p − 1 × p − 2 x p −3 x &c xq x q − 1 x q − 2 x 9-3 × &c × m 8xm-9x &c continued to m places. rxr-xr-xr-3x &c.x&e. continued to p, q, r, &c. plaçes refpectively, which is a more regular and fimple Series than them in the laft Parag. CHAP CHA P. II. Of the Combination of QUANTITIES. TH Definition. HE 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 Quantities. SCHOLIUM. From the Nature of Alternations and Combinations rightly confidered and compared, it will appear, That the Number of Alternations of m things abcde &c. different from each other, taken a by n, is equal to the Number of Combinations of the faid m things taken n by n, multiplied by the Number of Alternations of a things different from each other, taken n by n: But the Number of Alternations of the faid m things taken n by n, is, by the Coroll. in the last Chap. mx m—1 x m−2x m-3x &c. continued to places; and the Number of Alternations of n things different from each other taken a by n, is, by the faid Coroll. n= × n−1 × n−2 × n-3 × &c. continued to 2 places; therefore the Number of Combinations of the faid m things taken 2 by n is X &c. Each Series continued to 2 Placés or Terms. Example. Let it be required to find the Number of Combinations of the five things abcde different from each other, taken three by three. |