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seg, admit of an indefinite number of answers ; indeed, we may combine these seven sets of answers in as many ways as we wish, so that each combination shall produce a new answer. Thus, taking of the first set, and adding them to of the second set, we obtain the following answers.

231 pounds, at 6 cents. -
189 “ 6 8 6
189 « « 10 6
331

" 12 " If we take of each of the first three sets and add them, we

get

2331 pounds, at 6 cents.
163)

66 8
2331 6 6 10 6

210 " " 12 " Again, taking a seventh part of the sum of all the seren sets of answers, we get

2064 pounds, at 6 cents.
213 4 6 8
213 6 6 10 66
2062

12 " This method of combining can be varied in an infinite number of ways.

CHAPTER XIV.

PERMUTATION.

88. PERMUTATION means the number of transpositions which can be made with any number of individual things, all being taken at a time. Thus, by Permutation, we obtain the following results : 1) Perm. (abc) =

2) Perm. (aabb) =

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CASE I. : - When the individual things are all different, to find the number of permutations. From the above results, we deduce the following

RULE. Multiply continually the series of natural numbers, 1, 2, 3, 4, foco, until we have used as many factors as there are individual things.

. Exainples... 1. How many permutations can be made of 6 individual things, all different? Ans. 1X2 X3 X4 X5 X6=720. 2. How many different ways may 10 persons sit at table !

. Ans. 3628800. 3. How many permutations may be made with 15 individ. ual things? . . . .

: CASE II, When several of the individual things are of the same kind, we have this

RULE. I. Find the number of permutations, by Case I., on the sup. position that all the individual things are different.,

II. Find, by the same Case, the number of permutations, which can be made of as many individual things as there are terms of any particular sort ; do the same for all the terms which are repeated, then take the continued product of all the partial permutations, and divide the total number of permulations, as found by the first part of this rule, by this continued product.

. Examples. .. 1. How many permutations can be made of the letters A A BBCC? ::

SOLUTION.

. . : By Case I., we find the total punber of permutations of six things to be 1x2x3 X4 X5 X6. We also find the number of permutations of two things to be 1 X2; but since there are three letters which are repeated, each twice, our continued product of these partial permutations is 1 X2X1X2X1 X 2. Thomam 1X2 X3 X4 X5 X6

which, by canceling, redu. 1x2x1x2x1x2" ces to 90, for the number of permutations sought.

2. How many numbers, of nine figures each, can be expressed by the use of four figure ones, three figure twos, and two figure threes? 1 X2 X3 X4 X5 X6X7X8X9-1960

1X2 X3 X4X1X2X3X1 X 2 3. How many ways may the letters of the words HIGHER ARITHMETIC, be permutated ? Ans. 72648576000.

4. In how many different ways may three maple trees, five ash trees, and two elm trees, be set out in a single row!

COMBINATION,

89. COMBINATION means the number of ways in which a certain number of things can be combined, by taking a given number at a time. When two things are combined at a time, the combinations are said to be of the second class; when three are taken at a time, the combinations are then of the third class, and so on, for higher classes.

CASE 1. When the combinations allow of a repetition of the same in. dividual thing, we find the following results.

COMBINATIONS WITH REPETITIONS.

1) Comb. (a, b, c, d) of the second Class. aa, ab, ac, ad, bb, bc, bd, cc, cd, dd.

2) Comb. (a, b, c, d, e) of the second Class. aa, ab, ac, ad, ae, bb, bc, bd, be, cc, cd, ce, dd, de, ee.

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