Hence all numbers, from 1 to 15 inclusive, must be capable of being expressed in this system by 1 repeated not more than 4 times. But all the possible values of 1 in these 4 places are 1, 2, 4, or 8; hence all numbers, from 1 to 15 inclusive, are made up of some of the combinations of 1, 2, 4, and 8. Hence four weights of 1, 2, 4, and 8 ounces, will be sufficient to make up any number of ounces from 1 to 15 inclusive. This curious proposition may be easily extended. EXERCISES. 1. If the first digit in any number is divisible by 2, so is the number. 2. Any number having 5 or 0 in its place of units, is divisible by 5. 3. If a number is divided by 3, the remainder is the same as when the sum of its digits is divided by 3. 4. A number is divisible by 4 when the number composed of its first two digits is so; or when the sum of the first digit and twice the second is so. 5. A number is divisible by 8 when the number composed of its first three digits is so; or when the sum of the first digit, twice the second, and four times the third, is so. 6. If the sum of the odd digits and r times the even digits of a number (r being the base) is divisible by r+ 1, or r- 1, or 2-1, the number itself is so. 7. If the sum of the even digits of a number, expressed in the common system, be added to the number, and the sum of the odd digits be subtracted from it, the resulting number is divisible by 11. 8. In what system is the value of the number denoted by 231 equal to that of 66 in the decimal system? PERMUTATIONS. (455.) The various orders in which objects are capable of being arranged in succession, is called their permutations." Thus, the letters a, b, c, when taken in pairs, will form six permutations ab, ba, ac, ca, bc, cb; and when the three are taken, they will also form six, namely, abc, acb, bac, bea, cab, cba. THEOREMS. (456.) The number of permutations that can be formed with n letters, taken two and two, is n (n − 1)" Let a, b, c, ben letters, then a may be placed before each of the remaining letters, which are n -1 in number, and thus n 1 permutations are formed, in which a stands first. So b may be placed before each of the other letters; and thus n- - 1 permutations are formed, in which 6 stands first. The same may be said of all the letters which are n in number. Hence there are n - 1 permutations, repeated n times, or altogether, there are n (n-1) permutations. (457.) The number of permutations that can be formed with n letters, taken three and three, is n (n-1) (n — 2)1 ... ... ... For any one of the permutations, taken two and two, can be placed before each of the remaining letters, which are n 2 in number; and thus n-2 permutations, taken three and three, are formed. Thus, if a, b, c, d, e, be the n letters, then the permutation ab may be placed before each of the letters c, d, e, which are n-2 in number; and thus are formed n-2 permutations, taken three and three, namely, abc, abd, abe,... So the permutation ac being placed before each of the letters b, d, e, will form n— 2 permutations, taken three and three. The same may be proved of all the other permutations, taken two and two; and thus all the possible permutations, taken three and three, will be formed. It might be said that the first permutation, taken alone, namely, ab, may not only be placed before e forming the permutation abc, but that it might also be placed after e, so as to form another permutation cab; but this last permutation is otherwise formed, namely, by placing the permutation ca before b; so that to form all the permutations, taken three and three, it is merely necessary to place each of the permutations, taken two and two, before each of the remaining n-2 letters. Hence the whole number of permutations, taken three and three, will be equal to the number of them taken two and two repeated n-2 times, or n (n-1) (n-2) (458.) It may be similarly proved, that the number of permutations of n letters, taken four and four together, is n (n-1) (n-2) (n-3), by placing each of the permutations, taken three and three, before each of the remaining letters, which are n -3 in number." By proceeding in this manner, the following general proposition is arrived at by induction, namely:: (459.) The number of permutations of n letters, taken r and r together, is n (n − 1) (n − 2) ... (n—r—1); and hence ; The number of permutations of n letters, when all of them are taken, is n (n-1) (n − 2) 3 × 2 × 1, or 1 x 2 x 3 x (n − 2) (n − 1) n ... For in this case rn, and n-r ... the preceding factor is n-n-2=2; the one preceding this last is n— n 3= 3; and so on." (460.) The number of permutations of n letters, taken n -1 and n 1 together, is the same as when they are all taken;" or = 2 × 3 × 4 x · (n − 2) (n − 1) n For in this case rn ∙1, and r—1=n— ... ·2 = 2, n — r — EXAMPLES. -2; and 2=n 1. In how many different orders can five persons sit on a form? The number of permutations of 5 objects taken altogether is 1×2× ... (n − 1) n=1×2×3×4×5=120 2. How many signals may be made with 4 flags ? 3 by 3 is n (n-1) (n-2)=4x3x2 4 by 4 is n (n-1) (n − 2) (n —3)=4x3x2x1 = 24 EXERCISES. 1. How many permutations of four notes each, sounded successively, can be formed with thes even musical notes of one octave? Ans. 840. 2. In how many different ways can the seven prismatic colours be arranged? Ans. 5040. 3. In how many different ways can six letters be arranged when taken singly, two by two, three by three, and so on, till they are all taken together? Ans. 1956. COMBINATIONS. (461.) The different collections that can be formed by any number of objects, without regarding the order in which they are arranged, is called their combinations." Although several permutations may consist of the same objects, every two combinations must consist of different objects. THEOREMS. (462.) The number of combinations of n objects, taken For the number of permutations, two and two is n (n-1), and for each combination there are two permutations; as, for example, the combination ab affords two permutations ab, ba; therefore the number of combinations is permutations, or they are = n (n-1) 1 x 2 of the (463.) The number of combinations of n objects, taken n(n-1) (n-2), three and three, is 1 x 2 x 3 For the number of permutations, taken three and three, is n (n. 1) (n-2), and each combination as abc affords 6 or 2 × 3 permutations; hence the number of combinan (n-1) (n-2) tions = 1 x 2 x 3 (464.) Generally, the number of combinations of n objects, taken m and m together, is For the number of permutations, taken m and m toge bination of m objects affords 1x2x3.....(m– 1)m permutations; hence the number of combinations (465.) When all the quantities are taken together, there is evidently only one combination,-as appears also from the consideration that m is then = n, and the last n (n-1) (n-2).... 3 x 2 x 1 formula becomes = 1 EXAMPLE. 1. How many products can be formed with 5 different 1. How many different tints of colour can be formed by mixing the seven prismatic colours always in the same proportion, and in every variety of ways ? Ans. 120, |