342. By the preceding method we are enabled to discover the approximate value of a fraction expressed in large numbers, and this principle has some important applications, particularly in Astronomy. Ex. 4. The ratio of the circumference of a circle to its diamter is 3.1415926. Find approximate values for this ratio. Ans. 10, 113. 333 355 Ex. 5. The length of the tropical year is 365d. 5h. 48m. 48s. Find approximate values for the ratio of 5h. 48m. 48s. to 24 hours. Ex. 6. In 87969 years the earth makes 277287 conjunctions with Mercury. Find approximate values for the ratio of 87969 to 277287. Ex. 7. In 57551 years the earth makes 36000 conjunctions with Venus. Find approximate values for the ratio of 57551 to 36000. Ans. 8, 135. 음, Ex. 8. In 295306 years the moon makes 3652422 synodical revolutions. Find an approximate value for the ratio of 295306 to 3652422. 19 Ans. 235 Ex. 9. One French metre is equal to 3.2809 English feet. Find approximate values for the ratio of a metre to a foot. Ex. 10. One French kilogramme is equal to 2.2046 pounds avoirdupois. Find approximate values for the ratio of a kilogramme to a pound. Ex. 11. One French litre is equal to 0.2201 English gallons. Find approximate values for the ratio of a litre to a gallon. PERMUTATIONS AND COMBINATIONS. 343. The different orders in which things can be arranged are called their permutations. In forming permutations, all of the things or a part only may be taken at a time. Thus the permutations of the three letters a, b, c, taken all together, are abc, acb, bac, bca, cab, cba. The permutations of the same letters taken two at a time are ab, ac, ba, bc, ca, cb. The permutations of the same letters taken one at a time are a, b, c. 344. The number of permutations of n things taken m at a time is equal to the continued product of the natural series of numbers from n down to n-m+1. Suppose the things to be n letters, a, b, c, d . . . . . The number of permutations of n letters, taken singly or one at a time, is evidently equal to the number of letters, or to n. If we wish to form all the permutations of n letters taken two at a time, we must write after each letter each of the n-1 remaining letters. We shall thus obtain n (n-1) permu tations. If we wish to form all the permutations of n letters taken three at a time, we must write after each of the permutations of n letters taken two at a time each of the n-2 remaining letters. We shall thus obtain n (n−1)(n-2) permutations. In the same manner we shall find that the number of permutations of n letters taken four at a time is n(n−1)(n−2)(n—3). Hence we may conclude that the number of permutations of n letters taken m at a time is n(n-1)(n-2) (n−3).... (n−m+1). 345. The number of permutations of n things taken all together is equal to the continued product of the natural series of numbers from 1 to n. If we suppose that each permutation comprehends all the n letters; that is, if m=n, the preceding formula becomes n(n-1)(n-2).... 3×2×1; or, inverting the order of the factors, 1.2.3.4.... (n-1)n, which expresses the number of permutations of n things taken all together. For the sake of brevity, 1.2.3.4.... (n-1)n is often de noted by n; that is, in denotes the product of the natural numbers from 1 to n inclusive. 346. The combinations of things are the different collections which can be formed out of them without regarding the order in which the things are placed. Thus the three letters a, b, c, taken all together, form but one combination, abc. Taken two and two, they form three combinations, ab, ac, bc. 347. The number of combinations of n things, taken m at a time, is equal to the continued product of the natural series of numbers from n down to n―m+1 divided by the continued product of the natural series of numbers from 1 to m. The number of combinations of n letters taken separately, or one at a time, is evidently n. The number of combinations of n letters taken two at a n (n−1) time is 1.2 For the number of permutations of n letters taken two at a time is n (n-1), and there are two permutations (ab, ba) corresponding to one combination of two letters; therefore the number of combinations will be found by dividing the number of permutations by 2. The number of combinations of n letters taken three at a n (n−1)(n—2) time is 1.2.3 For the number of permutations of n letters taken three at a time is n(n-1)(n−2), and there are 1.2.3 permutations for one combination of these letters; therefore the number of combinations will be found by dividing the number of permutations by 1.2.3. In the same manner we shall find the number of combina tions of n letters taken m at a time to be n(n−1)(n−2) . . . . (n−m+1) 1.2.3.... m EXAMPLES. 1. How many different permutations may be formed of 8 letters taken 5 at a time? Ans. 8.7.6.5.4-6720. 2. How many different permutations may be formed of the 26 letters of the alphabet taken 4 at a time? Ans. 358800. 3. How many different permutations may be formed of 12 letters taken 6 at a time? Ans. 665280. 4., How many different permutations may be formed of 8 things taken all together? Ans. 1.2.3.4.5.6.7.8=40320. 5. How many different permutations may be made of the letters in the word Roma taken all together? 6. How many different permutations may be made of the letters in the word virtue taken all together? 7. What is the number of different arrangements which can be formed of 12 persons at a dinner-table? Ans. 479001600. 8. How many different combinations may be formed of 6 letters taken 3 at a time? 9. How many different combinations may be formed of 8 letters taken 4 at a time? Ans. 70. 10. How many different combinations may be formed of 10 letters taken 6 at a time? Ans. 210. 11. A telegraph has m arms, and each arm is capable of n distinct positions; find the total number of signals which can be made with the telegraph. 12. How many different numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, each of these digits occurring once, and only once, in each number? CHAPTER XVIII. BINOMIAL THEOREM. 348. The binomial theorem, or binomial formula, is a formula discovered by Newton, by means of which we may obtain any power of a binomial x+a, without obtaining the preceding powers. 349. By actual multiplication, we find the successive powers of x+a to be as follows: (x+a)2=x2+2ax+a2, found in the same manner, are as fol (x− a)2= x2-2ax+a2, (x —α)3=x3-3ax2+3a2x—a3, (x—a)1=x1—4ax3+6a2x2 —4a3x+aa, (x—α)3=x3 — 5αx2+10а2x3-10a3x2+5aax—a3. - On comparing the powers of x+a with those of x-a, we perceive that they only differ in the signs of certain terms. In the powers of x+a, all the terms are positive. In the powers of x-a, the terms containing the odd powers of a have the sign minus, while the terms containing the even powers have the sign plus. The reason of this is obvious; for, since a is the only negative term of the root, the terms of the power can only be rendered negative by a. A term which contains the factor -a an even number of times will therefore be positive; if it contain it an odd number of times it must be negative. Hence it appears that it is only necessary to seek for a method of obtaining the powers of x+a, for these will become the powers. of x-a by simply changing the signs of the alternate terms. |