185. The Permutations of any number of quantities signify the changes which these quantities may undergo with respect to their order.

Thus, if we take the quantities a, b, c; then, a b c, a c b, b a c, b c a, ca b, c ba, are the permutations of these three quantities taken all together; ab, ac, ba, bc, ca, cb, are the permutations of these quantities taken two and two; a, b, c, are the permutation of these quantities taken singly, or one and one, &c.

The problem which we propose to resolve is,

186. To find the number of the permutations of ʼn quantities, taken P together.

and P

The number of the permutation of these n quantities taken singly, or one and one, is manifestly n.

The number of the permutations of these n quantities, taken two and two together, will be n (n-1). For since there are n quantities,

If we remove a there will remain (n - 1) quantities,

Writing a before each of these (ʼn — 1) quantities, we shall have

That is, (n

1) permutations of the n quantities taken two and two, in which a stands first. Reasoning in the same manner for b, we shall have (n − 1) permutations of the n quantities taken two and two, in which b stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be

--

1) for the n,

1)

The number of the permutations of n quantities taken three and three together is n (n-1) (n-2). For since there are n quantities, if we remove a there will remain (n 1) quantities; but, by the last case, writing (n number of the permutations of (n − 1) quantities taken two and two is (n (n- 2); writing a before each of these (n - 1) (ʼn — 2) permutations, we shall have (n-1) (n · 2) permutations of the n quantities taken three and three, in which a stands first. Reasoning in the same manner for b, we shall have

(n

1) (n

2) permutations of the n quantities taken three and three in which b stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be

In like manner we can prove that the number of permutations of ʼn quantities taken four and four will be

Upon examining the above results, we readily perceive that a certain relation exists between the numerical part of the expressions, and the class of permutations to which they correspond.

Thus the number of permutations of ʼn quantities, taken two and two, is

n (n

− 1) which may be written under the form n (n − 2 + 1)

Taken three and three, it is

n (n − 1) (n − 2) which may be written under the form n (n − 1) (n − 3 + 1)

Taken four and four, it is

n (n − 1) (n — 2) (n — 3) which may be written under the form n (n − 1)

(n-2) (n-4+ 1)

Hence from analogy we may conclude, that the number of permutations of n things, taken Р and p together, will be

n (n. 1) (n 2) (n 3)

In order to demonstrate this, we shall employ the same species of proof already exemplified in (Arts. 39 and 89), and show that, if the above law be assumed to hold good for any one class of permutations, it must necessarily hold good for the class next superior.

Let us suppose, then, that the expression for the number of the permutations of n quantities taken (p-1) and (p

It is required to prove that the expression for the number of the permutations of n quantities, taken Р and p together, will be

r (n

· 1) (n − 2) (n — 3)............

........

..(n − p + 1)

Remove a one of the n quantities a, b, c, d............k, then by the expression (A), writing (n-1) for n, the number of the permutations of the (n - 1) quantities b, c, d......... .k, taken (p-1) and (p - - 1, will be

(n-1) (n-2) (n-3)............. {(n - 1) — (p −1) + 1}

permutations, we shall have (n - 1) (n − 2) (n — 3).........................(n.

tations of the n quantities, in which a stands first. Reasoning in the same manner

185. The Permutations of any number of quantities signify the changes which these quantities may undergo with respect to their order.

Thus, if we take the quantities a, b, c; then, a b c, a c b, b a c, b c a, ca b, c ba, are the permutations of these three quantities taken all together; ab, ac, ba, bc, ca, cb, are the permutations of these quantities taken two and two; a, b, c, are the permutation of these quantities taken singly, or one and one, &c.

The problem which we propose to resolve is,

186. To find the number of the permutations of n quantities, taken P and Р together.

Let a, b, c, d,

.........

k, be the n quantities.

The number of the permutation of these n quantities taken singly, or one and one, is manifestly n.

The number of the permutations of these n quantities, taken two and two together, will be n (n - 1). For since there are n quantities,

Writing a before each of these (n - 1) quantities, we shall have

That is, (n 1) permutations of the n quantities taken two and two, in which a stands first. Reasoning in the same manner for b, we shall have (n − 1) permutations of the n quantities taken two and two, in which 6 stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be

-

n,

the

The number of the permutations of n quantities taken three and three together is n (n − 1) (n-2). For since there are » quantities, if we remove a there will remain (n 1) quantities; but, by the last case, writing (n 1) for number of the permutations of (n − 1) quantities taken two and two is (n (n-2); writing a before each of these (n have (n − 1) (n — 2) permutations of the in which a stands first. Reasoning in the

·1) (n ·

2) permutations, we shall

n quantities taken three and three, same manner for b, we shall have

(n

1) (n

2) permutations of the n quantities taken three and three in which b stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be

In like manner we can prove that the number of permutations of n quantities taken four and four will be

Upon examining the above results, we readily perceive that a certain relation exists between the numerical part of the expressions, and the class of permutations to which they correspond.

Thus the number of permutations of n quantities, taken two and two, is

n (n − 1) which may be written under the form n (n − 2 + 1)

Taken three and three, it is

n (n − 1) (n − 2) which may be written under the form n (n − 1) (n − 3 + 1)

3) which may be written under the form ʼn (n − 1)

-

Hence from analogy we may conclude, that the number of permutations of n things, taken Р and p together, will be

In order to demonstrate this, we shall employ the same species of proof already exemplified in (Arts. 39 and 89), and show that, if the above law be assumed to hold good for any one class of permutations, it must necessarily hold good for the class next superior.

Let us suppose, then, that the expression for the number of the permutations of n quantities taken (p — 1) and (p

-

1) together is

It is required to prove that the expression for the number of the permutations of n quantities, taken P and p together, will be

Remove a one of the n quantities a, b, C, d.......

(A), writing (n - 1) for n, the number of the permutations of the (n - 1) quantities b, c, d............k, taken (p − 1) and (p — 1, will be

Or,

(n-1) (n-2) (n-3)................ {(n - 1) — (p −1)+1}

for b, we shall have (n

1) (n — 2) (n —3)................................ (n —p+1) permutations of the n quantities in which 6 stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be

n (n-1) (n-2) (n-3)....(n — p + 1)...(1)

Hence it appears, that, if the above law of formation hold good for any one class of permutations, it must hold good for the class next superior; but it has been proved to hold good when p = 2, or for the permutations of n quantities taken two and two, hence it must hold good when p 3, or for the permutation of n quantities taken three and three, .. it must hold good when and so on. The law is, therefore, general.

p = 4,

Example.

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

Here

n = 8, p = 5, n — · p + ) = 4 hence the above formula.

· 2) · · · · · (n − p + 1) = 8 × 7 × 6 × 5 × 4 = 6720

n (n-1) (n

the number required.

.....

Which expresses the number of the permutations of n quantities taken all together.*

Example.

Required the number of the permutations of the eight letters, a, b, c, d, v, f, g, h.

Here n = 8, hence the above formula (2) in this case becomes,

1.2 3 4 5.6 7 8 40320

the number required.

188. The number of the permutations of n quantities, supposing them all different from each other, we have found to be

Many writers on Algebra confine the term permutations to this class where the quantities are taken all together, and give the title of arrangements, or variations to the groupes of the n quantities when taken two and two, three and three, four and four, &c. The introduction of these additional designations appears unnecessary, but in using the word permutations absolutely, we must always be understood to mean those represented by formula (2), unless the contrary be specified.