would mix with grain at 35 cts. and at 56 cts.; how much of each of the two last must he take to form a compound worth 50 cts. a bushel, which shall fill a garner 10 ft. long, 5 ft. 4 in. wide, and 5 ft. high.

Ans. 69 bush. at 35 cts., and 106 at 56 cts.

6. How many bushels of grain at 40, 50, 60, and 80 cts. a bushel must we have to form a compound worth 56 cts. a bushel, which shall just fill the above garner?

Ans. 10317 at 40 cts., 17% @ 50, 25 @ 60, and 69, @80 cts.

Permutation.

441. The changes which can be made in the order of any given number of things, the whole being disposed in regular succession after each other, and each being found in every arrangement only once, are called permutations.

Suppose we have two things b and c : it is plain that they are susceptible of the two permutations be and cb, which we may express by 1 X 2. If we have three things a, b, c: as any one of them may stand first, each remaining two being susceptible of two permutations, it is plain that we shall have 3 times 2, or 6 permutations, namely, abc, acb, bac, bca, cab, cba, which number is expressed by 1 × 2 × 3, or 1.2.3.

Again, if we have four things: as any three of the four are susceptible of 6 permutations, and as each may be placed first in its turn, it is evident that the number of permutations of which they are susceptible is 4 times 6 24, or 1.2.3.4.

Now it is easy to see that if there be 5 things, the number of changes in their order will be 5 times 24, or 1.2.3.4.5; and that in general, whatever is the number of things, the number of permutations of which they are susceptible is expressed by the continual multiplication of the natural numbers 1.2. 3.4.5, &c., ending with the number corresponding to the given number of things.

Supplement to Permutation.

442. When, of the things which are given, several are alike, as a, a, a, b, it is plain that as those things which are alike are not susceptible of permutation, the number of permutations of which the like things would have been susceptible, had they all differed from each other, shows how many times less the number of permutations of which the given things are susceptible now is, than it would have been had they been all differ

ent.

We therefore operate thus: Find the number of permutations of which the given things would have been susceptible had they been all different, as usual; and divide this number by the product of the numbers which show the number of permutations of which all the like things would have been susceptible, if different.

Thus, for the four things a, a, a, b, we have

1.2.3.4

4, 1.2.3 the number of permutations of which they are susceptible. These are in effect, aaab, aaba, abaa, baaa, and are evidently all of which they are susceptible, seeing that b is put in every place from the first to the last inclusive, and that the letters a, a, a, are not susceptible of permutation.

For the permutations of the ten letters aabbbbcccd, we have 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 (1.2) (1.2.3.4) (1.2.3)

Examples.

= 12600.

1. How many changes can be made in the order of the 5 vowels?

Ans. 120. 2. In how many ways can the seven notes of music be played? Ans. 5040. 3. How many different numbers, each consisting of 9 figures, can be expressed with the 9 digits? Ans. 362880.

4. How many numbers, each consisting of ten figures, can be expressed with the figures 7, 1, 2, 3, 8, 3, 2, 9, 2, 6? Ans. 302400. 5. In how many ways can a vignette be formed, which shall consist of 2 violets, 3 primroses, 4 tulips, and 5 roses ? Ans. 2522520.

Arrangement.

443. When we have a number m of things, and are required to find how many changes can be made in their order when they are taken 2 by 2, 3 by 3, 4 by 4, &c., n by n, n being always assumed less than m, these changes are called arrange

ments.

Suppose we have 5 things a, b, c, d, e, and we wish to know how many changes can be made in their order by taking them 2 by 2. As any one can precede each of the others, it is evident that 4 arrangements will thus result from each, and consequently that 5 things taken 2 by 2 are susceptible of 5 times 4, or 20 arrangements.

Now if we have any number m of things, it is equally plain that by placing any one before each of the others, the number of arrangements thus formed will be (m-1), and that by thus placing the whole of the units in m, each in its turn, we shall have m times (m - 1), or m(m —1) for the whole number of arrangements of which m things are susceptible when taken 2 by 2.

Again, if we take 5 things 3 by 3, we easily see that, before the whole of the arrangements obtained in placing them 2 by 2, we can place each one of the remaining 3, and thus we shall have 3 times 20, or 60 arrangements. Also, when the things in a number m are arranged 2 by 2, we can, before every arrangement, place each of those which remain, the number of which is (m2); consequently, as the number of arrangements 2 by 2 is m(m 1), the number of arrangements 3 by 3 must be 2) times m (m (m - 1); or, inverting the order of the multiplication, m (m1) (m—2).

Hence we see, that the number of arrangements of which a number m of things is susceptible, when taken 2 by 2, is m(m-1), and when taken 3 by 3, is m(m-1) (m—2).

444. Now, if we take the given number of things n by n, n-being less than m, it is easy from the above reasoning to infer that the number of arrangements, n by n, of which m things are susceptible, is expressed by the continual multiplication of the decreasing series m(m-1) (m2) (m—3) (m 4), &c., in the last term of which the negative number is equal to (n-1); and as this number is to be subtracted from m, the last term in the series will be m (n − 1), or (m · n + 1).

The general formula, therefore, for finding the number of arrangements of m things, taken n by n, is m(m — 1) (m — 2) (m3), &c...... (m — n + 1).*

Examples.

1. Suppose we demand how many different numbers, each consisting of 4 figures, can be formed with the. six digits 1, 2, 3, 4, 5, 6.

The negative number in the second term being 1, and that in the last (n-1), it is plain that the number of terms succeeding m is (n-1); consequently, the whole number of terms will always equal n. Wherefore, to find the number of arrangements, we multiply a series, beginning with the given number of things, and decreasing by a unit, till the number of terms in the series is equal to the number of things taken at a time.

The last term of the series will, according to the formula, be (6 − 4 + 1) = 3; we shall, therefore, have

6 × 5 × 4 × 3 =360, the required number.

2. How many different whole numbers, each consisting of 5 figures, can be formed with the 9 digits? Ans. 15120. 3. How many different whole numbers, each consisting of 7 figures, can be expressed with the nine digits?

Ans. 181440. 4. If from a regiment consisting of 10 companies, we take 6 companies at a time, in how many ways can the 6 companies take up their order of march? Ans. 151200.

Combination.

445. Of the arrangements of which m things are susceptible when taken 2 by 2, 3 by 3, &c., n by n, n being less than m, those which differ from each other by at least one of the things of which they are composed, are called combinations. Thus, the letters a, b, c, taken 2 by 2, are susceptible of the 6 arrangements ab, ac, ba, bc, ca, cb, of which only 3, ab, ac, and bc, are combinations.

Now, as the combinations of m letters, taken n by n, all differ from each other by at least one of the letters in n, it is plain that all the permutations of which the n letters of each combination are susceptible are so many new arrangements of the m letters taken n by n, and comprise all the arrangements of which the m letters, so taken, are susceptible. Hence the whole number of arrangements, of which m letters, taken n by n, are susceptible, is the product of the combinations of the same m letters, taken n by n, multiplied by the number of permutations of which n letters are susceptible. Wherefore, calling the whole number of arrangements of the m letters A, the combinations C, and the permutations of n letters P, it is plain,

A

that A PC, and consequently C= That is to

=

P

number of combinations of m letters, taken n by n, is equal to the whole number of their arrangements when so taken, divided by the number of permutations of which the n letters are susceptible. But (444) A m(m — 1) (m 2) (m 3) &c......(m − n + 1) and (441) P=1.2.3.4.5, &c...... 1) X n. Wherefore, the expression C

X (n

A

be

P

comes C

m(m − 1) (m — 2) (m − 3)..............(m — n + 1)

—

4, &c....... X (n-1)n

Examples.

1. Suppose that having 12 different kinds of flowers, we would know the whole number of nosegays, each consisting of 7 different kinds, that we can with the 12 kinds form, so that no two nosegays may consist of exactly the same kinds of flowers. It is evident that, to solve this question, we have only to find the combinations of 12 things taken 7 by 7. Wherefore, 12×11×10×9× 8 × 7 × 6 by the above formula, we have I 2 3 4 5 6 7 792, the required number.

2. With 9 different kinds of fruit, how many persons can we regale, so that each may partake of 4 kinds, and no two persons exactly the same kinds?

Ans. 126.

Ans. 924.

3. How many different numbers can we form by multiplying 6 at a time of the first 12 prime numbers? 4. From 40 men, how many times could a commander select 20, without again choosing the same 20?

Ans. 137846528820 times.

Supplement to Combination.

446. If we have two sets of series of different letters, which we would combine 2 and 2, it is evident that in combining a letter of the first set with each of those in the second, we shall have as many combinations, 2 and 2, as there are letters in the second set; wherefore, in thus combining every letter of the first set, it is plain that all the combinations, 2 and 2, of which the two sets are capable, in choosing one from each, is signified by their product. Again, if we have a third set, and we would combine the whole 3 by 3, by taking one from each, it is plain that in combining each letter of the third set with each of the combinations, 2 and 2, of the two first, we shall have the number of combinations, 3 and 3, which is signified by the product of the first and second series. Wherefore, the whole number of combinations of which 3 sets are capable, when taken 3 by 3, is signified by the product of the three sets; and, pursuing the analogy, all the combinations which can be made from any number of sets of different things, by taking, to form each combination, one from each set, is expressed by the continual multiplication of the numbers which show the number of things in each set.