2. Take the series 1×2×3×4, &c. up to the number of the given things of the first sort, and the series, 1×2×3×4, &c. up to the number of the given things of the second sort, &c. 3. Divide the product of all the terms of the first series by the joint product of all the terms of the remaining ones, and the quotient will be the answer required. EXAMPLES. 1. How many variations may be made of the letters in the word Zaphnathpaaneah ? 1×2×3×4×5×6×7×8×9×10×11×12×13×14×15 ( = 120 = = number of letters in the word) = 1307674368000. 1×2×3×4×5 ( = number of as) 1x2 (= number of ps) 1 (= number of is) 1X2X3 (= number of hs) 1x2 (number of ns) 2×6×1×2×120=2880)1307674368000(454058600 Ans. = 6 = 2 2. How many different numbers can be made of the following figures, 1223334444 ? PROBLEM IV. Ans. 12600. To find the number of combinations of any given number of things, all different from one another, taken any given number at a time. RULE.* 1. Take the series 1, 2, 3, 4, &c. up to the number to be taken af a time, and find the product of all the terms. 2. Take a series of as many terms, decreasing by 1, from the given number, out of which the election is to be made, and find the product of all the terms. 3. Divide the last product by the former, and the quotient will be the number sought. EXAMPLES. Any 3 quantities, a, b, c, all different from each other, admit of 6 variations; but if the quantities are all alike, or, a b c become aaa, then the 6 variations will be, 1X2X3 1. Again, if two quantities reduced to 1, which may be expressed by out of three are alike, or abc become aar; then the 6 variations will be reduced to 1X2X3 1X2X3 terms; where m is the number of given quantities, and time. X; &c. to ? those to be taken at a Note. In any given number of quantities, the number of Combinations increafes gradually till you come about the mean numbers, and then gradually decreases. If the number of quantities be even, half the number of places will fhew the greatest number of Combinations, that can be made of thofe quantities; but if odd, then those two numbers which are the middle, and whofe fum is equal to the giv en number of quantities, will fhew the greatest number of Combinations. EXAMPLES. 1. How many combinations may be made of 7 letters out of 12? 1x2×3×4×5×6×7 (= the number to be taken at a time)=5040. 12x11x10x9x8x7x6( same number from 12)=3991680. I 5040)3991680(792 Ans. 2. How many combinations can be made of 6 letters out of the 24 letters of the alphabet? Ans. 134596. 3. A general was asked by his king what reward he should confer on him for his services; the general only required a penny for every file, of 10 men in a file, which he could make out of a company of 90 men: What did it amount to? Ans. £.23836022841 7s. 1134d. 4. A farmer bargained with a gentleman for a dozen sheep, (at 2 dollars per head) which were to be picked out of 2 dozen; but being long in choosing them, the gentleman told him that if he would give him a cent for every different dozen which might be chosen out of the two dozen, he should have the whole, to which the farmer readily agreed: Pray what did they cost him? Ans. D.27041 56c. 5. How many locks, whose wards differ, may be unlocked with a key of 6 several wards? Ans. 636 of which may have one single ward, 15 double wards, 20 triple wards, 15 four wards, 6 five wards, and 1 lock 6 wards. To find the number of combinations of any given number of things, by taking any given number at a time; in which there are several things of one sort, several of another, &c. 1 RULE. Find the number of different forms, which the things, to be taken at a time, will admit of, in the following manner : 1. Place the things so that the greatest indices may be first, and the rest in order. 2. Begin with the first letter, and join it to the second, third, fourth, &c. to the last. 3. Join the second letter to the third, fourth, &c. to the last; and so on till they are all done, always rejecting such combinations as `have occurred before; and this will give the combination of all the twos. 4. Join the first letter to every one of the twos; then join the second, third, &c. as before; and it will give the combinations of all the threes. 5. Proceed 5. Proceed in the same manner to get the combinations of all the fours, fives, &c. and you will at last get all the several forms of combination, and the number in each form. 6. Having found the number of combinations in each form, add them all together, and the sum will be the number required. 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 4 of these quantities. 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, anabbc? No. 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. 9x9x9x9x9=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, 6x6x6x6=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 5x5x5x5=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 cbosen out of three companies, consisting of 5 men, 7 women and 9 children? Ans, 315. MISCELLANEOUS 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 of 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(-038462% 78 220 208 120 104 160 156 8 This multiplied by 4 (the numerator) is 15384118 Which annexed to the quotient·03846 is·0384615384, And 0384615384×8 and annexed to the last product ·0384615384307692307613, &c. 4. 2. Reduce 246* 10 246)1.000000(004065 and 0040650 x 10 = 0040650 100 246 246 and this annexed to the quotient is 004065406501, and this multiplied by the given numerator, 5, is 0203270325246. 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 tre numbers, whose difference is small, compared with the two numbers. FOR THE DUPLICATE RATIO. RULE.-Assume two numbers, whose difference is small; subtra& half their difference from the least, and add it to the greatest, and the two numbers, thus found, will be in the same proportion nearty as the squares of the assumed numbers. EXAMPLE. Let the assumed numbers be 10 and 11; Then 11-10=1. 5-9.5 and 11+5=11·5. Proof, As 102: 112 :: 9.5: 11.5 nearly. FOR A TRIPLICATE RATIO. 10 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 |