posite the ingredient with which that value is connected, and the difference (or the sum of the differences, if there be more than one) opposite each ingredient will be the quantity of that ingredient required. EXAMPLE FOR THE BOARD. How much sugar, at 5cts., 7cts., 8cts., 10cts., and 12cts., must be mixed together, that the mixture may be worth 9cts. a pound? 1st Ans. 4lb. at 5, 1 lb. at 7, 4lb. at 8, 77b. at 10, 5lb. at 12. 2d Ans. 3lb. at 5, 3lb. at 7, 1lb. | 3d Ans. 11b. at 5, 31b. at 7, 3lb. at 8, We may obtain as many answers as there are different ways of connecting the numbers above, with those below the average. To prove the rule correct, let us examine the second of the above answers. If we were mixing sugars at 5 and 12cts. to sell the mixture at 9cts., we should gain 4cts. on every pound of the former, and lose 3cts. on every pound of the latter. Then, on 3lb. of the former we should gain 12cts. and on 4lb. of the latter we should lose 12cts.; therefore, if we mix these quantities, we shall neither gain nor lose by selling the mixture at 9cts. In the same way it may be shown that 3lb. at 7cts. and 2lb. at 12cts., 1lb. at 8cts. and 1lb. at 10cts. may be sold at the average of 9cts., and the same reasoning will prove the truth of each of the other answers. 1. How much tea at 50cts. 75cts. 90cts. and $1.00, must be taken to form a mixture worth 80cts. 2. A jeweller has gold of 16, 17, 18, 20, 22, and 24 carats fine. What proportion of each must he take to make a mixture 21 carats fine? Pure gold is 24 carats fine, or 24. 3. How much grain, at 50cts. 75cts. $1.00, and $1.10 per bushel, will make a mixture worth 90cts. a bushel 4. How much water must be mixed with wine at $1.50 and $2.00 a gallon, to make the whole worth $1.00 per gallon? 5. What quantity of raisins, at 10cts. 18cts. and 20cts. per lb. must be mixed together, to fill a cask containing 150lb. and to be worth 19cts. a lb.? [After obtaining the proportions by Alligation, find the exact quantities by Fellowship.] 6. It is required to mix sugar at 7cts. 8cts. 10cts, and 12cts. per lb. in such manner as to form a mixture of 2cwt. 3qr., worth 11cts. per lb. 7. Mix tobacco at 8cts. 10cts. 12cts. and 16cts. so as to make 100 pounds worth 11cts. a pound. EXAMPLE FOR THE BOARD. A farmer wishes to mix 10 bushels of barley at 50cts., 4 bushels of oats at 45cts., and 16 bushels of rye at 75cts. with wheat at $1.25 and corn at 90cts. a bushel, so that the mixture may be worth $1.00 per bushel. 1.00 { 6290 25 We may regard the limited quantities as a single ingredient of 30 bushels, worth 62cts. a bushel. Proceeding in the usual way, we find that 25 bushels at 62cts., 25 at 90cts., and 48 at $1.25, would give us 1.25- 138+10 a mixture of the desired average value. But as we have 30 bushels at 62cts., we must take 3 or g of these proportionate quantities, and we have 30 bushels at 90cts. and 573bu. at $1.25, for the Answer. 30 8. How much water must be mixed with 40 gallons of syrup, at 50cts. a gallon, to make the whole worth 37 cts. a gallon? 9. A grocer has 10 gallons of wine at 75cts., 12gal. at 90cts., and 8gal. at $1.00, with which he would mix brandy at $1.25, and water, so as to make a mixture worth 95cts. a gallon. How much of each must he take? 10. If a cubic inch of gold weighs 11.11oz. and a cubic inch of silver 6.04oz., what quantity is contained in a mass of 63oz., of which 1 cubic inch weighs 8.35oz.? 11. How much molasses, at 50 cents, and water, must be mixed with 15 gallons at 37 cents, 28 gallons at 25 cents, and 19 gallons at 33 cents, to make a mixture worth 31 cents a gallon? 12. A grocer has an order for 150lb. of tea, at 90 cents per lb., but having none at that price, he would mix some at 75 cents, some at 87 cents, and some at $1.00 per pound. How much of each sort must he take? CHAPTER XIV. PERMUTATION AND COMBINATION. PERMUTATION shows the number of changes that can be made in the order of a given number of things. PROBLEM I. To find the number of changes that can be made of any given number of things, all different from each other. How many changes may be made in the position of 4 persons at table? If there were but two persons, a and b, they could sit in but two positions, ab and ba. If there were three, the third could sit at the head, in the middle, or at the foot, in each of the two changes, and there could then be 1×2×3=6 changes. If there were 4, the fourth could sit as the 1st, 2d, 3d, or 4th, in each of these 6 changes, and there would then be 1×2×3×4=24 changes. RULE. Multiply together the series of numbers 1, 2, 3, &c., up to the given number, and the product will be the number sought. 1. How many variations may be made in the order of the 9 digits? 2. How many changes may be made in the position of the letters of the alphabet? 3. How long a time will be required for 8 persons to seat themselves at table in every possible order, if they eat 3 meals a day? PROBLEM II. Any number of different things being given, to find how many changes can be made out of them by taking a given number of the things at a time. If we have five things, each one of the 5 may he placed before each of the others, and we thus have 5×4 permutations of 2 out of 5. If we take 3 at a time, the third thing may be placed as 1st, 2d, and 3d, in each of these permutations, and we have 5×4×3 permutations of 3 out of 5. For a similar reason we have 5×4×3×2 permutations of 4 out of 5, &c. RULE. Take a series of numbers, commencing with the number of things given, and decreasing by 1, until the number of terms is equal to the number of things to be taken at a time: the product of all the terms will be the answer required. 4. How many changes can be rung with 8 bells, taking 5 at a time? 5. How many numbers of 4 different figures each, can be expressed by the 9 digits? 6. In how many different ways may 10 letters of the alphabet be arranged? PROBLEM III. To find the number of permutations in any given number of things, among which there are several of a kind. How many permutations can be made of the letters in the word terrier ? If the letters were all different, the permutations, according to Problem I. would be 1×2×3×4×5×6x7=5040. But the permutations of the three r's would, if they were all different, be 1×2×3, which could be combined with each of the other changes; the number must therefore be divided by 1x2x3. For the same reason it must also be divided by 1×2, on account of the 2 e's. Then the true number sought is 1×2×3×4×5×6×7 RULE. =420. Take the natural series, from 1 up to the number of things of the first kind, and the same series up to the number of things of each succeeding kind, and form the continued product of all the series. By the continued product divide the number of permutations of which the given things would be capable if they were all different, and the quotient will be the number sought. 7. How many changes can be made in the order of the letters in the word Philadelphia? 8. How many different numbers can be made, that will employ all the figures in the number 119089907343? 9. How many permutations can be made with the letters in the word Cincinnati ? COMBINATION shows in how many ways a less number of things may be chosen from a greater. If we have ten articles, each may be combined with every one of the nine remaining ones, and therefore we may have 10x9 permutations of 2 out of 10. But each combination will evi dently be repeated; thus, we shall have ob and ba, ac and ca, &c. 10×9 Therefore, the number of combinations will be. 2 If now we add an eleventh article, each of the eleven may be joined to each of the combinations of the remaining ten, and we shall have 11×10×9 permutations. But each combination will 1X2 be three times repeated; thus we shall have abc, bac, and cab; abd, bad, and dab, &c. The number of combinations of 3 out of 11 will therefore be lowing 11 X 10×9 1x2x3 RULE. Hence we obtain the fol Write for a numerator the descending series, commencing with the number from which the combinations are to be made, and for a denominator the ascending series, commencing with 1, giving to each series as many terms as are equivalent to the number in one combination. Cancel the like factors in the numerator and denominator, and divide. 10. How many combinations of 4 letters, can be made from the alphabet? 11. How many combinations of 7 can be made from 18 apples? 12. How many ranks of 10 men, may be made in a company of 80? 13. How many locks of different wards, may be unlocked with a key of 6 wards? [Find the number of combinations of 1, 2, 3, 4, 5, and 6 in 6, and the sum of all the combinations will be the number required.] CHAPTER XV. INVOLUTION. INVOLUTION is the repeated multiplication of a number by itself. The product obtained by Involution is called a power. The root is the number involved, or the first power. If the root be multiplied by itself, or employed twice as a factor, the product is the second power. If the root is |