EXAMPLE. Let the things proposed be aaabbe: It is required to find the number of combinations of every 2, of every 3, and of every of these quantities. Combinations at large. aa,aa,ab,ab,ac aa,ab,ab,ac ab,ab,ac 4 bb,bc bc a3 Ans. 5 combinations of every 2; 6 of every 3, and 5 of every 4 quantities. PROBLEM VI. To find the changes of any given number of things, taken a given. number at a time; in which there are several given things of one sort, several of another, &c. RULE. 1. Find all the different forms of combination of all the given things, taken, as many at a time, as in the question, by Problem 5. 2. Find the number of changes in any form, (by Problem 3,) and multiply it by the number of combinations in that form. 3. Do the same for every distinct form, and the sum of all the products will give the whole number of changes required. EXAMPLE. How many changes can be made of every 4 letters out of these 6, aaabbc? No. of forms. Comb. Changes. 1×2×3×4=24 -= 4 1x2x3 = 6 2X 4 8 Therefore, 1X 6= 6 38 = number of changes required. PROBLEM VII. To find the compositions of any number, in an equal number of sets, the things being all different. RULE. Multiply the number of things in every set continually together, and the product will be the answer required, EXAMPLES. 1. Suppose there are five companies, each consisting of 9 men; it is required to find how many ways 5 men may be chosen, one out of each company? Multiply 9 into itself continually, as many times as there are companies. 9×9×9×9×9=59049 different ways, Ans. 2. How many changes are there in throwing 4 dice? As a die has 6 sides, multiply 6 into itself four times continually. 6×6×6×6=1296 changes, Ans. 3. Suppose a man undertakes to throw an ace at one throw with 4 dice, what is the probability of his effecting it? First, 6×6×6×6=1296 different ways with and without the ace. Then, if we exclude the ace side of the die, there will be 5 sides left; and 5×5×5X5 625 ways without the ace; therefore there are 1296-625-671 ways, wherein one or more of them may turn up an ace; and the probability that he will do it, as 671 to 625, Ans. 4. In how many ways may.a man, a woman and a child be chosen out of three companies, consisting of 5 men, 7 women and 9 children? Ans. 315. MISCELLANEOUS MATTERS. A short method of reducing a Vulgar Fraction, into its equivalent Decimal, by Multiplication. RULE. Divide unity or 1 by the denominator, till the remainder is a single figure, 10, 100, &c. if convenient, then multiply the whole quotient, including the remainder after division, by the remainder (which is now the numerator, and the divisor, the denominator) and annex the product to the quotient, then multiply the quotient, thus increased by the last numerator, and annex the product to the increased quotient; and thus it may be reduced to what exactness you please. But if the numerator of the given fraction exceed 1, you must finally multiply the last product by the said numerator. EXAMPLES. 1. Reduce to its equivalent decimal. 26)1.00(03846 78 This multiplied by 4 (the numerator) is 1538418=;'; Which annexed to the quotient 03846 is 0384615384, 220 And 0384615384,8×8 and annexed to the last product= 0384615334307692307613, &c. 208 120 104 160 156 4 2. Reducer. 246)1.000000(004065% and 0040650x100040650 and this annexed to the quotient is 0040654065018, and this multiplied by the given numerator, 5, is 02032703252. For any number of pounds, avoirdupois, under 28, multiply the decimal 00892857 by the given number of pounds, which generally gives the decimal true to the sixth place. A short method of finding the duplicate, triplicate, &c. Ratio of any. two numbers, whose difference is small, compared with the two numbers. FOR THE DUPLICATE RATIO. RULE. Assume two numbers, whose difference is small; subtract half their difference from the least, and add it to the greatest, and the two numbers, thus found, will be in the same proportion nearly as the squares of the assumed numbers. EXAMPLE. Let the assumed numbers be 10 and 11; Then 11-10=1. 10-5-9.5 and 11+5=11.5. 4 Proof, As 102: 112 :: 9.5: 11·5 nearly. FOR A TRIPLICATE RATIO. RULE. Subtract the difference of the assumed numbers from the least, and add it to the greatest; and the numbers, thus obtained, will be in the same proportion nearly as the cubes of the assumed numbers. Let the numbers be 164 and 165: Then 165-164-1. 164-1 163 and 165+1=166. Proof, As 1643: 1653 :: 163: 166 nearly. 1 For a quadruplicate proportion subtract, and add once and a half the difference, and so on, for each higher power, increasing the number to be subtracted and added by ·5. To reduce à Ratio, consisting of large numbers, to its least terms, and very nearly of the same value. RULE. 1. Divide the greater of the terms by the less, and the least divisor by the remainder, and so on continually, till nothing remain, in the same manner as we get the greatest common measure for reducing a vulgar fraction: This will give a number of ratios, from which we can choose one, that will suit our purpose. 2. Place the first quotient under unit for the first ratio; multiply that by the next quotient, adding nothing to the numerator, and 1 to the product of the denominator, for a new denominator, and it will give a second ratio, nearer than the first: Then, multiply the last ratio by the next quotient, adding the preceding ratio, and so on, continually till you have gone through. EXAMPLES. 1. Sir Isaac Newton has demonstrated, in his Principia, that the velocity of a comet, moving in a parabola, is to that of a planet, moving in a circular orb, at the same distance from the sun, as√2 to 1. Let this be taken for an example. √2=1·4142; those motions, then, are as 14142 to 1 ; or as 14142 to 10000 ? The late Professor Winthrop chose 7 to 5 for a proportion. 2. Geometers have found the proportion of the circumference of a circle to its diameter, to be as 3-1416 to 1: Let this ratio be 3. The area of a circle is to its circumscribing square, as 7854 to 1, very nearly: Let this be reduced. 7854)10000(1 7854 2146)7854(3 1416)2146(1 3X1+1= 4 -third. Therefore, as 14: 11: the square of the diameter of a circle to its area. To estimate the Distance of Objects on level ground, or at sea, having only the height given. RULE. 1. To the earth's diameter, (viz. 42056462 feet,) add the height of the eye, and multiply the sum by that height, then the square root of the product is the distance, at which an object on the surface of the earth or water, can be seen by an eye so elevated. 2. As objects are seen in a straight line, and that line is a tangent to the earth's surface; therefore, To find the distance of two elevated objects, when the right line joining them touches the earth's surface |