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Ex. 1. What annuity, continued for 4 year.-;, at 6 per cent, compound interest, is now worth $ 207.90?

Ans. $ 6O.

The present value represented by Opsratio*. the debt, divided by the present

$ 207.90 -r- 3.465 = $ 60. worth of $ 1 for the given time and

rate, gives the annuity required.

Rule. Divide the yiren 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 nmount of an annuity, the time and rate, are given, the annuity mny be found by dividing ihe given amount by the amount of $ 1 for the given lime 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 number of changes that can be made in the arrangement of any given number of things.

580. Combination shows how often a less number of things combined can be taken out of a greater, without respect to their order.

581. To find the number of changes that can be made with any given number of things, taken all at once.

Ex. 1. How man.y 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

1X2X3 = 6. possible permutations that can be made of the first two letters of the alphabet; as, a b and b a. 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.

582i 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.

Operation. It is evident that

8 X (8 - 1) X (8 - 2) X (8 - 3) -J « £ ££J* = 8X7X6X5 = 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 8x7x6 permutations. In like manner for 4 out of 8, we obtain 8x7x6x5 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 trill be the answer required,

2. How many changes can be rung with 4 bells out of 6?

Ans. 360.

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.

Operation. 1 We find the number of permuta

4 X $ X 2 tions which may be made by taking 3

-— -— = ^ = 4. out of 4 letters, as in Art. 582, and the

1X2X3 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, $•<:., up to the less numher 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 die 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.

586^ Questions in which the required numbers, or their parts, or their multiples, are increased or diminished by some given number which is no known part or multiple of the required number, or when a power or root of the required number is either directly or indirectly contained in the result given in the question, may be solved by two assumptions.

Note. — When the answer is obtained by means of a single assumption, the process is called Position; and when obtained by means of two assumptions, it is called Double Position.

Analysis by Position affords often a very compendious method of working questions, whose solution otherwise, except by algebra, would be lengthy and difficult. It is even found very useful in shortening some of the processes of algebra.

587. When the answer may be obtained by means of a single assumption or supposition.

Ex. 1. A schoolmaster, being asked how many scholars he had, replied, that if he had as many more as he now has, and half as many more, he should have 200. Of how many scholars did his school consist? Ans. 80 scholars.

opKKAtio.i. J5V having as many more, and

Assumed number, 6 0 half as many more, he would have As many more, 6O llad 2itimes the original number; Tr ,„ o „ therefore, the required number

Half as many more, 3 0 must be as nianv a's 200 contains

15 0 times 2£, or 80. The same result A A A c A A may De obtained, as in the opera

150:200::60:8 0, Ans. tion, thus: We assume the number

of scholars to be 60; if to 60 as many move, and half as many more, are added, the sum is 150. As this result has the same ratio to the result in the question as the sup. posed number has to the number required, we find the answer by a proportion.

Rule. Assume am/ convenient number, and proceed with it according to the nature of the question. Then, if the result be either too much or too little, as the result found is to the result given, so will be the number assumed to the number required.

2. A person, after spending £ and £ of his money, had $ 60 left; what had he at first? Ans. $144.

3. What number is that, which, being increased by and £ of itself, equals 125?

4. A's age is double that of B, and B's is triple that of C, and the sum of all their ages is 140. What is each person's age? Ans. A's 84, B's 42, C's 14 years.

5. A person lent a sum of money at 6 per cent., and at the end of 10 years received the amount, $ 560. What was the sum lent? Ans. $ 350.

588. When two or more assumptions or suppositions are required in finding the answer.

Ex. 1. A lady purchased a piece of silk at 80 cents per yard, and lining for it at 30 cents per yard; the silk and lining contained 15 yards, and the price of the whole was $ 7. How many yards were there of each?

Ans. 5 yards of silk; 10 yards of lining.

OPERATION.

Assume 6 yards of silk, $ 4.80 Assume 4 yards of silk, $ 3.20 Lining would be Oyd., 2.70 Lining would be 11yd., 3.30

Their sum, $ 7.50 Their sum, $ 6.50

Sum in the question, 7.00 Sum in the question, 7.00 First error, —(- § 0.50 Second error, —$0.50

.50 -|_ .50 : 6 — 4 :: .50 : 1; 6 yards — 1 yard r= 5 yards; 15 yards — 5 yards = 10 yards.

Since the silk and lining contain 15 yards, cost $ 7.00, the average price per yard is 46$ cents; and 80 cents — 4(i| cents = 881 cents; 46$ cents — 30 cents = 10§ cents; and the quantity of lining will therefore be to that of the silk as 33J is to 16J, or as to 2 is to 1. Hence, if the given number of yards, 15, be divided into 3 parts, two of those parts, or 10 yards, will be for the lining, and the other part, or 5 yards, will be for the silk.

The same result is obtained as in the operation, thus: We assume the quantity of silk to be 6 yards; then the lining would be 9 yards. We find the cost of each of these at the prices given, and, adding, have as the sum of their costs $ 7.50. This is a result too large by $0.50, when compared with the sum in question. By assuming the quantity of the silk to be 4 yards, and proceeding in a like manner, we obtain for the cost of the silk and the lining $6.50, a sum too small by $0.50. The first error, arising from a result too large, is marked with the sign plus (-f-), and the second error, arising from a result too small, is marked with the sign minus (—). Then, since jne of the results is too large, and the other too small, we then say, as the sum of the errors, or .50 -4- .50, is to the difference of the assumptions, or 6 — 4, so is .50 the less error, to 1, the correction; ,^^ich, being added to 4 (yards), the sum, 5, expresses the number of yards of silk; and consequently that of the lining must be 10 yards.

Rule. Assume two different numbers, perform on them separately the opefations indicated in the question, and note the ERRORS of the re

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