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 9ly. 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 let8 X (8 — 1) X (8 — 2) X (8 - 3) 2) X (8 — 3) ters may be arranged 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 f out of 4 letters, as in Art. 582, and the 1 X 2 X 8 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, &c., 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. OPERATION. 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. By having as many more, and Assumed number, 60 halt as many more, he would have As many more, 60 had 24 times the original number; Half as many more, 3 0 therefore, the required number must be as many as 200 contains 150 times 21, or 80. The same result may be obtained, as in the opera150 : 200 :: 60:80, Ans. tion, thus: We assume the number of scholars to be 60; if to 60 as many more, 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 any 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 I of his money, had $ 60 left; what had he at first ? Ans. $144. 3. What number is that, which, being increased by 1, }, 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. sum lent? 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 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. Their sum, First error, Assume 6 yards of silk, $ 4.80 Assume 4 yards of silk, $ 3.20 Lining would be 9yd., 2.70 Lining would be 1lyd., 3.30 $ 7.50 Their sum, $ 6.50 Sum in the question, 7.00 Sum in the question, 7.00 + $0.50 Second error, $0.50 .50 +.50 : 6 — 4 :: .50 :1; 6 yards 1 yard = 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 463 cents; and 80 cents 46 cents = 33 cents; 463 cents 30 cents - 163 cents; and the quantity of lining will therefore be to that of the silk as 33} is to 163, 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 (+), and the second error, arising from a result too small, is marked with the sign minus (-). Then, since occ of the results is too large, and the other too small, we then say, as the sum of the errors, or .50 + .50, is to the difference of the assumptions, or 6 — 4, so is .50 the less error, to 1, the correction; Trich, 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 operations indicated in the question, and note the ERRORS of the res sults. Then, as the difference of the errors, if the results be both too great or both too small, or as the sum of the errors, if one result be too great and the other too small, is to the difference of the assumed numbers, so is either error to the correction to be applied to the number which produced that error. NOTE 1. · The rule usually given fails in an important class of questions; but the rule here given, if not the simplest in the resolution of some questions, has the advantage of being applicable in every case. NOTE 2. – In relation to all questions which in algebra would be resolved by equations of the first degree, the differences between the true and the assumed numbers are proportional to the differences between the result given in the question and the results arising from the assumed numbers. But the principle does not hold exactly in relation to other questions; hence, when applied to them, the above rule, or any other of the kind that can be given, will only produce approximations to the true results. In which case the assumed numbers should be taken as nearly true as possible. Then, to approximate more nearly to the required number, assume for a second operation the number found by the first, and that one of the first two assumptions which was nearer the true answer, or any other number that may appear to be still nearer to it. In this way, by repeating the operation as often as may be necessary, the true results may be approximated to any assigned degree of accuracy. This process is sometimes applied with advantage in extracting the higher roots, when approximate results, differing but slightly from entire correctness, will answer. 2. A and B invested equal sums in trade; A gained a sum equal to of his stock, and B lost $ 225; then A's money was double that of B's. What did each invest? Ans. $ 600. 3.- A person, being asked the age of each of his sons, replied, that his eldest son was 4 years older than the second, his second 4 years older than the third, his third 4 years older than the fourth, or youngest, and his youngest half the age of the oldest. What was the age of each of his sons ? Ans. 12, 16, 20, and 24 years. 4. A gentleman has two horses, and a saddle worth $ 50. Now if the saddle be put on the first horse, it will make his value double that of the second horse; but if it be put on the second, it will make his value triple that of the first. What was the value of each horse ? Ans. The first, $ 30; second, $ 40. 5. A gentleman was asked the time of day, and replied, that of the time past from noon was equal to go of the time to midnight. What was the time? Ans. 12 minutes past 3. 6. A and B have the same income. A saves ik of his, but B, by spending $ 100 per annum more than A, at the end of 10 years finds himself $ 600 in debt. What was their income? Ans. $ 480. |