1

2+1

3+1

2+1

2+•

(255.) When a fraction has been transformed into a continued fraction, its approximate value may be found by taking a few of the first terms of the continued fraction.

proximation; and three terms would give a still more accurate

value.

(256.) By this 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 diameter is 3.1415926. Find approximate values for this

ratio.

22 333 355

Ans.

7106' 113°

Ex. 5. In 87969 years, the Earth makes 277287 conjunctions with Mercury. Find approximate values for the frac

tion

87969

277287*

Ans.

1 6 7 13 33 3' 19' 22' 41' 104'

Ex. 6. In 57551 years, the Earth makes 36000 conjunctions

57551

with Venus. Find approximate values for the fraction 36000*

Ex. 7. In 295306 years, the Moon makes 3652422 synodical revolutions. Find an approximate value of the fraction. 295306

3652422

19

Ans.

235

THEORY OF PERMUTATIONS AND COMBINATIONS.

(257.) The different orders in which quantities may be arranged are called their Permutations. Thus,

a, b, c,

a, c, b,

the permutations of the three letters a, b, c, taken all b, a, c, together, are

b, c, a,

c, a, b,

c, b, a.

(a, b,

a, c,

The permutations of the same letters taken two and b, a,

α,

b,

The permutations of the same letters taken sugly, or one by one, are

C.

(258.) To find the number of permutations of n letters, taken m and m together.

Let

a, b, c, d..... k, be the n letters.

The number of permutations of n letters taken singly, or one by one, is evidently equal to the number of letters, or to n.

The number of permutations of n letters taken two and two is n(n-1). For if we reserve one of the letters, as a, there will remain n-1 letters,

b, c, d,.... k.

Writing a before each of these letters, we shall have ab, ac, ad . . . . . ak;

that is, we obtain n-1 permutations of the n letters taken two and two, in which a stands first. Proceeding in the same manner with b, we shall find n-1 permutations of the n letters taken two and two, in which b stands first; and so for each of the n letters. Hence the whole number of permutations will be

n(n-1).

The number of permutations of n letters taken three and three together is

n(n-1) (n-2).

For if we reserve one of the letters, as a, there will remain n-1 letters. Now we have found the number of permutations of n letters taken two and two to be n(n-1). Hence the permutations of n-1 letters taken two and two must be

(n-1) (n-2).

Writing a before each of these permutations, we shall have (n-1) (n-2) permutations of the n letters taken three and three, in which a stands first. Proceeding in the same manner with b, we shall find (n-1) (n-2) permutations of the n letters taken three and three, in which b stands first; and so for each of the n letters. Hence the whole number of permutations will be

n(n-1) (n-2).

In like manner, we can prove that the number of permutations of n letters taken four and four is

n(n−1) (n−2) (n−3).

When the letters are taken two and two, the last factor in ne formula representing the number of permutations is n-1. When the letters are taken three and three, the last factor is n-2. When the letters are taken four and four, the last

factor is n-3.

Hence, when the letters are taken m and m together, the last factor will be n-(m-1) or n―m+1; and the number of permutations of n letters taken m and m together will accordingly be

n(n-1) (n-2) (n-3).....(n-m+1).

EXAMPLES.

Ex. 1. Required the number of permutations of the 8 letters a, b, c, d, e, f, g, h, taken 5 and 5 together.

Here

n=8, m=5, n−m+1=4,

8.7.6.5.4-6720, Ans.

and the above formula becomes

Ex. 2. Required the number of permutations of the 26 letters of the alphabet, taken 4 and 4 together.

Ans. 358800.

Ex. 3. Required the number of permutations of 12 letters, taken 6 and 6 together.

Ans. 665280.

(259.) 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) ..... 2×1;

or, inverting the order of the factors,

1.2.3.4..... (n-1)n;

which expresses the number of permutations of n letters taken all together.

Ex. 1. Required the number of changes which can be rung upon 8 bells.

According to the preceding formula, we have

1.2.3.4.5.6.7.8=40320, Ans. Ex. 2. How many permutations may be formed from the letters of the word Roma?

Ex. 3. What is the number of permutations which may be formed from the letters composing the word "virtue ?”

Ex. 4. What is the number of different arrangements which can be made of 12 persons at a dinner-table?

Ans. 479001600.

(260.) The combinations of any number of quantities signify the different collections which may be formed of these quantities, without regard to the order of their arrangement.

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.

(261.) To find the number of combinations of n letters, taken m and m together.

The number of combinations of n letters taken separately, or one by one, is evidently n.

is

The number of combinations of n letters taken two and two, n(n-1)

1.2

For the number of permutations of n letters taken two and two 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 and n(n-1) (n-2)

three together, is

1.2.3

For the number of permutations of n letters taken three and three, is n(n-1) (n−2); and there are 1.2.3 permutations for one combination of three 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 combinations of n letters, taken m and m together, to be

n(n−1) (n−2)......... (n−m+1)

Ex 1. Required the number of combinations of six letters taken three and three together.

Ex. 2. Required the number of combinations of 8 letters taken 4 and 4.

Ans. 70.

Ex. 3. Required the number of combinations of 10 letters taken 6 and 6.

Ans. 210

The following table, which is computed by the preceding formula, shows the number of combinations of 1, 2, 3, 4, &c let