x2-x=35; which resolved, we have x=35.25+.5, whence x−6=√35.25+5.5, and .x+6=√/35.25+6.5; the numbers therefore are .43717, 6.43717,and 12.43717, nearly. 2. An artist proposed to work as many days at 3 shillings per day, as he had shillings in his pocket; at the end of the time having received his hire, and spent nothing, he finds himself worth 44 shillings; what sum did he begin with ?. Let x=his number of shillings at first, whence also x=the number of days he worked: we have therefore here given the first term x, the common difference 3, and the number of terms x+1, in an arithmetical progression, to find the last term; now by theor. 1. (z=a+n−1.d, or) 44=x+x+1−1×3, that is, 4x= 44, whence x=11 shillings the sum he began with. 3. To find three numbers in arithmetical progression, such, that their sum may be 12, and the sum of their squares 56? Let x=the common difference, 3 s=(12) the sum, then will s=the middle number, s-x=the less extreme, and s+x=the greater extreme, also let a=56; then by the problem, (s—x)2+ s2+s+x}2=) 3s2+2x2=a, whence 2x2-a-3s2, and x= a- -3 $2 2 56-48 = =2; therefore s=4,s-x=2, and s+x=6, that is, 2, 4, and 6, are the numbers required. 4. To find four numbers in arithmetical progression, whereof the product of the extremes is 52, and that of the means 70? Let x=the less extreme, y=the common difference; then will x, x+y, x+2y, and x+3y, represent the progression. Let a= 52, b=70, then by the problem (x.x+3y=) x2+3xy=a, and (x+y.x+2y=) x2+3xy+2y2=b; from the latter equation subb-a tract the former, and 2 y2=b-a, whence y=√・ stitute this value for y in the first equation, and it becomes x2+9x =a; completing the square, &c. we obtain x=√✅at 2 wherefore 4, 7, 10, and 13, are the numbers required. =3; sub 81 9 :4: 5. The sum of six numbers in arithmetical progression is 48, and if the common difference d be multiplied into the less extreme, the product equals the number of terms; required the terms of the progression? 6x5 2 6 16 5 d=16, or (putting d 5 12 5 stitution, 48=6a+· -.d, that is, 6a+ 15 d=48; whence 2 a+ for a) 5 d+12=16 d, or d2——d= ; whence by completing the square, &c. d=2, therefore a= 3; consequently the numbers are 3, 5, 7, 9, 11, and 13. 6. The continual product of four numbers in arithmetical progression is 380, and the sum of their squares 214; what are the numbers ? Let p 880, s=214, 2x=the common difference, y—3x= the less extreme; then will y-3 x, y−x, y +- x, and y+3x=the terms of the progression: wherefore by the problem, y—3 x.y—x. y+x.y+3x=p, and y-3x2+y−x)2+y+x}2+y+3x]2=s; equations reduced, become y1—10 y2x2+9x1=p, and 4 y2+20 x2 =s; from the latter of these y2==-5xo, therefore y*= 5 8.x2 2 2 Ꭶ 4 these 16 +25x+; if these values be substituted for their equals in the = R2 4 or (putting a= and = ; then by completing the square, &c. x= =(by restoring the values of a and R) 14; whence y= (√—-—5x2=) 64: therefore y—3 x=2, y—x=5, y+x=8, and y+3x=11, the numbers required. 20. To find the number of permutations, which can be made with any number of given quantities. Def. The permutations of quantities are the different orders in which they can be arranged. Let a and b be two quantities; these will evidently admit of two permutations, viz. ab and ba, which number of permutations may be thus expressed, 1x2. Let a, b, and c, be three quantities; these admit of six permu→ tations, abc, bac, cab, acb, bca, and cba, viz. 1×2×3. Let a, b, c, and d, be four quantities; these admit of 24 permutations; thus, abcd bacd cabd dabc abdc badc cadb dacb adcb bdca cdba deba That is, 4 things admit of 1×2×3×4 permutations. And therefore n things admit of 1x2x3, &c. to n, permutations. EXAMPLES.-1. How many ways can the musical notes ut, re, mi, fa, sol, la, be sung? Ans. 1×2×3×4×5×6=720 ways. 2. How many changes can be rung on 12 bells? Answer, 479001600. 3. How many permutations can be made with the 24 letters of the alphabet ? 21. To find the number of combinations that can be made out of any given number of quantities. Def. The combinations of quantities, or things, is the taking a less collection out of a greater as often as it can be done, without regarding the order in which the quantities so taken are arranged. Thus, if a, b, and c, be three quantities, then ab, ac, and bc, are the combinations of these quantities, taken two and two: and here it is necessary to remark, that although ab and ba form two different permutations, yet they form but one combination; in the same manner ac and ca make but one combination, as also bc and cb. Let there be n things given, namely a, b, c, d, &c. (to n terms,) then if a be placed before each of the rest, n-1 permutations will be formed; if b be placed before each of the rest, n-1 permutations will in like manner be formed; and if c, d, e, &c. be placed respectively before each of the rest, n-1 permutations in each case will arise; consequently, if each of the n things be placed before all the rest, there will be formed in the whole n.n-1 permutations; that is, there can n.n-1 permutations be formed of n things taken two at a time. Hence, if instead of n we suppose n—1 things, b, c, d, e, &c. the number of permutations which these afford of the quantities taken two and two, will (by what has been shewn) be n-1.n-2; now if a be prefixed to each of these permutations, there will be n-1.n-2 permutations in which a stands first; in the same manner it appears, that there will be n—1.n-2 permutations in each case when b, c, d, e, &c. respectively stand first; and therefore when each of the n things have stood first, there will be formed in the whole n.n-1.n-2 permutations of n things taken three and three. By similar reasoning it appears that n things taken 4 at a time afford n.n― -1.n- -2.n-3 \permu tations. This being premised, we may readily obtain the number of combinations, each consisting of 2, 3, 4, 5, &c. to r things, which can be made out of any given number n; for it appears by the preceding problem, that 2 things admit of 2 permutations, but by the definition they admit of but 1 combination; and therefore any number of things taken 2 at a time, admit of half as many combinations as there are permutations; but the number of permutations in n things, taken two and two, has been shewn to be n.n-1; therefore the number of combinations in n things, If three things be taken at a time, then 6 permutations will arise from every 3 things so taken, and but 1 combination; and therefore any number of things taken 3 at a time, admit of one sixth as many combinations, as there are permutations; but the number of permutations in n things taken 3 at a time, has been shewn to be nn-1n-2; and therefore the number of combina tions in n things, taken 3 at a time, will be n.n-1.n-2 1.2.3 By similar reasoning it may be shewn, that the number of combinations in n things, taken EXAMPLES.-1. How many combinations can be made of 2 letters, out of 10? 2. How many combinations of 5 letters can be made out of the 24 letters of the alphabet? Here n=24, then n.n-1.n-2.n-3.n-4 1.2.3 4.5 =10626. Ans. 3. In a ship of war there are 40 officers, and the captain intends to invite 6 of them to dine with him every day; how many parties is it possible to make, so that the same 6 persons may not meet at his table twice? 22. To investigate the rules of simple interest. Def. 1. The sum lent is called the principal. 2. The money paid by the borrower to the lender for the use of the principal, is called interest. 3. The interest (or quantity of money to be paid) is previously agreed upon; that is, at a certain sum for the use of 1001. for a year: this is called the rate per cent. per annum o. Per cent. means by the hundred, and per annum, by the year; the term 5 per cent. per annum, means 5 pounds paid for the use of 100%. lent during the space of a year, &c. Various rates of interest have been given in this country for the use of |