for b, we shall have (n − 1) (n — 2) (n —3)............(n—p + 1) permutations of the n quantities in which 6 stands first, and so on for each of the " quantities in succession, hence the whole number of permutations will be Hence it appears, that, if the above law of formation hold good for any one class of permutations, it must hold good for the class next superior; but it has been proved to hold good when p = 2, or for the permutations of a quantities taken two and two, hence it must hold good when p = 3, or for the permuta tion of n quantities taken three and three, .. it must hold good when p = 4, and so on. The law is, therefore, general. Example. Required the number of the permtations of the eight letters, a, b, c, d, e, f. g, h, taken 5 and 5 together. Here n = 8, p = 5, n − p + 1 = 4 hence the above formula. n (n-1) (n-2). . . . . ( − p + 1) = 8 X 7 X 6 X5 X 4 = 6720 Which expresses the number of the permutations of n quantities taken all together.* Example. Required the number of the permutations of the eight letters, a, b, c, d f, g, h. Here n = 8, hence the above formula (2) in this case becomes, 188. The number of the permutations of n quantities, supposing them all different from each other, we have found to be Many writers on Algebra confine the term permutations to this class where the quantities are taken all together, and give the title of arrangements, or variations to the groupes of the n quantities when taken two and two, three and three, four and four, &c. The introduction of these additional designations appears unnecessary, but in using the word permutations absolutely, we must always be understood to mean those represented by formula (2), unless the contrary be specified. But if the same quantity be repeated a certain number of times, then it is ma ..ifest that a certain number of the above permutations will become identical. Thus, if one of the quantities be repeated a times, the number of identical permutations will be represented by 1. 2. 3........... ................., and hence, in order to obtain the number of permutations different from each other, we must divide (2) by 1. 2. 3............, and it will then become • If one of the quantities be repeated a times, and another of the quantities be repeated 3 times, then we must divide by 1 2............a X 1. 2............ß; and, in general, if among the n quantities there be a of one kind, ẞ of another kind, of another kind, and so on, the expression for the number of the permutations different from each other of these " quantities will be 1 2 3....... 1 2 3..... n 1 2......a X 1.2.........ß X 1. 2......y, &c. 1 2 3........... 1.2 3 4. 5 6 Example I. Required the numbers of the permutations of the letters in the word algebra. Here" = 7, and the letter a is repeated twice, hence formula (3) becomes 2 3. Example 2. Required the number of the permutations of the letters in the word ceifacaratadaddara. Here n = 18, a is repeated eight times, c twice, d thrice, r twice, hence the number sought will be · 1.2.3.4.5.6.7. 8 9. 10. 11. 12. 13. 14. 15. 16. 17. 18 1.2.3.4.5.6.7.8 x 1.2 x 1.2.3 × 1.2 ..(3) Example 3. Required the number of the permutations of the product a at full length. Heren = x + y + z, the letter a is repeated x times, the letter b, y times and the letter c, z times; the expression sought will therefore be = 6616209Eff L •• (x + y + z) 1 2. 3......x X 1. 2. 3......y × 1. 2, 3......z b c", written 189. The Combinations of any number of quantities, signify the different collec tions which may be formed of these quantities, without regard to the order in which they are arranged in each collection. Thus the quantities a, b, c, when taken all together, will form only one com. bination, abc; but will form six different permutations, abc, acb, bac, bca, cab, cba; taken two and two they will form the three combinations ab, ac, bc, and the six permutations ab. ba, ac, ca, bc, cb. The problem which we propose to resolve is, 190. To find the number of the combinations of u quantities, taken p and p together. Let the number of combinations required be x: Suppose these x combinations to be formed and to be written one after the other, in a horizontal line; write below the first of these all the permutations of the p letters which it contains, and since the number of these is 1.2.3......p (=y suppose), we shall have a vertical column consisting of y terms; the second term of the horizontal line will, in like manner, give another vertical column consisting of y terms, being all the permutations of the P letters which it contains, one at least of which is different from those in the combinations already treated of. The third combination will, in like manner, give y terms differing from all the others. We shall thus form a table consisting of z columns, each of which contains y terms; and on the whole xy results, which are evidently all the permutations of the n letters, taken p and p together, none being either omitted or repeated; we shall therefore have by formula (1), ...... (nCp) = .(n − p + 1) the expression required. Р Hence we perceive, that the number of the combinations of n quantities, taken and p together, is equal to the number of the permutations of n quantities, taken P and p together, divided by the number of the permutations of p quantitres taken all together. There is a species of notation employed to denote permutations and combinations, which is sometimes used with advantage from its conciseness. ........ and P, The number of the permutations of n quantities, taken P (n Pn) and *............. are represented by (nCp) and so on. It is manifest that the above proposition may be expressed accord ing to this notation by (n Pp) ..(4) P, METHOD OF UNDETERMINED COEFFICIENTS. 191. The method of undetermined coefficients is a method for the expan sion or development of algebraic functions into infinite series, arranged according to the ascending powers of one of the quantities considered as a variable. The principle employed in this method may be stated in the following THEOREM. If the series A+Bx+Cx2+Dx3+ &c., whether finite or infinite, be equal to the series A'+B'x+C'x2+D13+ &c., whatever be the value of x; then the coefficients of the like powers of a must be the same in each series; that is, A=A1, B=B', C=C', D=D', &c. ̧ For since A+B+Ca2+Da3+ &c. =A1+B1x+C1x2+D1a3+ &c. by transposition we have (A—A1)+(B-B1)x+(C—C1)x2+(D—D1)x3+. =0. Now, if all or any of these coefficients were not =0, the equation would determine particular values of x, and could only be true for such particular values, which is contrary to the hypothesis. Hence we must have A-A1=0, B-B1-0, C-C1=0, &c., and therefore A=A1, B=B1, C=C1, &c. (1.) Expand the fraction 1 Assume 1-2x+2a 1 1-2x+x2 =A+Bx+Cx2+ Dx3 + Ex1+ &c. 1 1-2x+x2 EXAMPLES. then, multiplying by 1-2x+x2, we have 1=A+ Bx+ €œ2 + Da3 + Ex1+ -2Ax-2Bx2-2Cx3-2Dx*— +Ax2+ Ba3+ Ca1+ hence, by the preceding theorem, we have A=1 B-2A=0 C-2B+ A=0 D-2C+B=0 E-2D+C=0 into an infinite series. x 2 =1 .. A= &c. . . Therefore =1+2x+3x2+4x3+5x1+6x2+ This example has been chosen to illustrate the method of expansion by undetermined coefficients; but the development can be obtained by division in the usual way, or by synthetic division, with more facility than by the principle here employed. (2.) Extract the square root of 1+x. Assume √1+x=A +Bx+Cx2+Dr3+ ..., and square Doth sides; +ACx2+ BC☛3+C2x1 + . +ADr3+BDr1+. .... hence, equating the coefficients of the like powers of x, we have By quadratics we find -13x+40=(x−5) (x-8); hence we may as and by the principle of undetermined coefficients we have A+B=3, and 8A+5B = 5; Note. The values of A and B might have been determined in the fol lowing manner: 3x-5 Since A B A(x−8)+B(x-5) .'. 3x—5— A(x—8) + B(x−5). Now this equation must subsist for every value of x; and, therefore, This method may frequently be employed with advantage, and will be found useful in the integration of rational fractions, when we come to treat of the Integral Calculus. |