OPERATION, years, at 6 years, at 5 The present worth re$3.465106 X 60 = $ 207.906+. quired evidently may be ob tained by finding the amount of the given annuity, by the last articles, and then finding in the usual way the present worth of that amount. A more expeditious method, however, is to find, in the table, the present worth of an annuity of $ 1 for the given time and rate, and take that sum as many times as there are dollars in the given annuity, as in the operation. RULE. — Multiply the present worth of an annuity of $ 1 for the given time and rate by the number denoting the given annuity. 2. What is the present worth of an annuity of $100, for 9 per Ans. $ 680.169. 3. What is the present worth of an annuity of $ 200, for 7 per Ans. $ 1157.27. 4. Required the present worth of an annuity of $ 500, to continue 40 years, at 7 per cent. 5. A gentleman wishes to purchase an annuity, which shall afford him, at 6 per cent. compound interest, $ 500 a year, for ten years. What sum must he deposit in the annuity office to produce it? Ans. $ 3680.04. 6. If a widow be entitled to $ 160 a year, payable semiannually, from a fund, for 8 years, what is its value at present, at 6 per cent. compound interest ? Ans. $ 1004.88. 576. To find the present worth of an annuity in perpetuity. Ex. 1. What is the present worth of a perpetual lease, which yields an income of $ 600, the rate of interest being that of 6 per cent. ? Ans. $ 10000. The question is evidently the same as one requiring what principal in one $ 600 +.06 $ 10000. year, at 6 per cent. interest, will yield $ 600. RULE. Divide the given annuity by the number denoting the interest of $ 1 for one year. NOTE. — When the annuity is payable quarterly, semiannually, or in any other periods less than a whole year, the annuity must be increased by the interest which may thus accrue on the parts of the annuity payable before the end of the year, before dividing by the interest of $ 1 for one year. 2. A ground rent in the city of Philadelphia yields an annual income of $ 963, at 6 per cent. interest. What is the value of the estate ? Ans. $ 16050. OPERATION. OPERATION. 3. What is the present value of a perpetual lease, yielding an income of $ 6335, interest being at 7 per cent. Ans. $ 90500. 4. What sum should be paid for a perpetual annuity of $ 1200, payable semi-annually, interest being at 5 per cent. ? Ans. $ 24000. 577. To find the present worth of an annuity in reversion. Ex. 1. What is the present worth of an annuity of $ 300, to commence in 3 years, and to continue 5 years, allowing compound interest at 6 per cent. ? Ans. $ 1061.03. The present worth $ 6.209744 $ 2.673012 $ 3.536732; of an annuity of $1, $ 3.53673 X 300 = $ 1061.03. at 6 per cent., com. mencing at once, and continuing till the termination of the annuity, or for 3 + 5 = 8 years, is $ 6.209744; and the present worth of the same annuity up to the time of the commencement of the reversion is $ 2.673012. The difference of these present worths multiplied by the number of dollars in the given annuity is the present worth of the reversion. RULE. — Find the present worth of an annuity of $ 1, commencing immediately and continuing till the reversion COMMENCES, and also till the reversion TERMINATES ; and multiply the difference of these present worths by the number of dollars in the given annuity. The result will be the present worth required. 2. The reversion of a lease of $ 350 per annum, to continue 11 years, which commences 9 years hence, is to be sold. What is its worth, allowing the purchaser 6 per cent. per annum for his ready money? Ans. $ 1633.70. 3. A father presents to his daughter, for 8 years, a rental of $ 70 per annum, payable yearly, and its reversion for the 12 years succeeding to his son. What is the present value of the gift to his son, allowing 4 per cent. compound interest? Ans. $ 480.03. 4. What is the present worth of the reversion of a perpetuity of $ 240 per annum, payable yearly, but not to come into possession till the expiration of 100 years, compound interest being allowed at 6 per Ans. $ 11.78. 578. To find the annuity, the present worth, time, and rate being given. OPERATION. Ex. 1. What annuity, continued for 4 years, at 6 per cent. compound interest, is now worth $ 207.90 ? Ans. $ 60. The present value represented by the debt, divided by the present $ 207.90 = 3.465 = $ 60. worth of $ 1 for the given time and rate, gives the annuity required. RULE. Divide the given present worth by the present worth of an annuity of $ 1 for the given time and rate, and the result will be the annuity required. NOTE. — When the amount of an annuity, the time and rate, are given, the annuity may be found by dividing the given amount by the amount of $1 for the given time and rate. 2. The present value of an annuity, to be continued 10 years, at 6 per cent. compound interest, payable annually, is $3680.04; required the annuity. Ans. $ 500, 3. An annuity, remaining unpaid for 9 years, at 5 per cent. compound interest, amounted to $882.125 ; what was the annuity ? 4. A yearly pension which has been forborne for 6 years, at 6 per cent., amounts to $ 279; what was the pension ? Ans. $ 40. PERMUTATIONS AND COMBINATIONS. 579. PERMUTATION is the process of finding the nuinber of changes that can be made in the arrangement of any given number of things. COMBINATION shows how often a less number of things combined can be taken out of a greater, without respect to their order. 580. 581. To find the number of changes that can be made with any given number of things, taken all at once. Ex. 1. How many changes of order do the first three letters of the alphabet admit of? Ans. 6. OPERATION. By trial we shall find that two are all the 1 X 2 X 3 = 6. possible permutations that can be made of the first two letters of the alphabet; as, ab and ba. If we take an additional letter, 6 are all the possible permutations; as, abc, acb, bca, bac, cab, cba. Now, the same result may be obtained, in the case of the two letters, by multiplying together the first two digits, and in case of the three letters by multiplying together the first three digits, as in the operation. RULE. Multiply together all the terms of the natural series of numbers, from 1 up to the given number, inclusive, and the product will be the number required. 2. How many changes may be rung on 6 bells ? Ans. 720 changes. 3. For how many days can 10 persons be placed in a different position at dinner ? 4. How many changes may be rung on 12 bells, and how long would they be in ringing, supposing 10 changes to be rung in one minute, and the year to consist of 365 days 5 hours and 49 minutes ? Ans. 479001600, and 91y. 26d. 22h. 41m. 5. How many changes do the letters of the alphabet admit of? Ans. 403291461126605635584000000. OPERATION. 582. To find how many changes may be made by taking each time any number of different things less than all. Ex. 1. How many sets of 4 letters each may be formed out of 8 different letters? Ans. 1680. each one of the 8 let – 2) X (8 — 3) ters may be arranged 8 X (8 — 1) X (8 — 2) X (8 — 3) = 8 X 7 X 6 X 5 = 1680. before each of the oth ers; therefore 2 out of the 8 letters admit of 8 x 7 permutations. By taking 3 out of the 8 letters, the third letter can be arranged as the first, second, and third, in each of the same permutations, giving 8 X 7 X 6 permutations. In like manner for 4 out of 8, we obtain 8 X 7 X 6 X 5 permutations. RULE. Take a series of numbers, beginning with the number of things given, and decreasing by 1, until the number of terms equals the number of things to be taken at a time, and the product of all the terms will be the answer required. 2. How many changes can be rung with 4 bells out of 6 ? Ans. 360. OPERATION. = 4. 3. How many words can be made out of the 26 letters of the alphabet, 6 being taken at once? Ans. 165765600. 583. To find the number of combinations that can be formed from a given number of different things, taken a given number at a time. Ex. 1. How many combinations can be made of 3 letters out of 4, the letters all being different ? Ans. 4. We find the number of permuta4 x 3 x 2 tions which may be made by taking 3 $ out of 4 letters, as in Art. 582, and the 1 X 2 X 3 number of permutations by the three letters all at once, as by Art. 581, and, dividing the first by the latter, obtain the number of combinations required. RULE. Take the series, 1, 2, 3, 4, 5, 8c., up to the less number of things, and find the product of the terms. Take also a series of numbers beginning with the greater number of things, and decreasing by 1 until the number of terms equals the less number of things, and find the product of the terms. The latter result divided by the former will give the number required. 2. How many combinations can be made of 7 letters out of 10, the letters all being different ? Ans. 120. 3. A successful general, being asked what reward would satisfy him for his services, demanded only a cent for every file of 10 men which he could make with a body of 100 men. What would his demand amount to? Ans. $ 173103094564.40. ANALYSIS BY POSITION. 584. ANALYSIS BY Position is the process of solving analytical questions, by assuming or supposing one or more numbers, and reasoning from them, operated upon as if they were the number or numbers required to be found. 585. Questions in which the required number is in any way increased or diminished in any given ratio, or in which it is multiplied or divided by any number, may be solved by means of a single assumption. |