### Part I. Section II. Chapter 11. “Of the Transposition of the Letters, on which the demonstration of the preceding Rule is founded.”

352 If we trace back the origin of the coefficients which we have been considering, we shall find, that each term is presented, as many times as it is possible to transpose the letters, of which that term is composed; or, to express the same thing differently, the coefficient of each term is equal to the number of transpositions which the letters composing that term admit of. In the second power, for example, the term $$ab$$ is taken twice, that is to say, its coefficient is 2; and in fact we may change the order of the letters which compose that term twice, since we may write $$ab$$ and $$ba$$. The term $$aa$$, on the contrary, is found only once, and here the order of the letters can undergo no change, or transposition. In the third power of $$a+b$$, the term $$aab$$ may be written in three different ways; thus, $$aab$$, $$aba$$, $$baa$$; the coefficient therefore is 3. In the fourth power, the term $$a^3b$$ or $$aaab$$ admits of four different arrangements, $$aaab$$, $$aaba$$, $$abaa$$, $$baaa$$; and consequently the coefficient is 4. The term $$aabb$$ admits of six transpositions, $$aabb$$, $$abba$$, $$baba$$, $$abab$$, $$bbaa$$, $$baab$$, and its coefficient is 6. It is the same in all other cases.

353 In fact, if we consider that the fourth power, for example, of any root consisting of more than two terms, as $$(a+b+c+d)^4$$, is found by the multiplication of the four factors,

$(a+b+c+d)(a+b+c+d)(a+b+c+d)(a+b+c+d),$

we readily see, that each letter of the first factor must be multiplied by each letter of the second, then by each letter of the third, and, lastly, by each letter of the fourth. So that every term is not only composed of four letters, but it also presents itself, or enters into the sum, as many times as those letters can be differently arranged with respect to each other; and hence arises its coefficient.

354 It is therefore of great importance to know, in how many different ways a given number of letters may be arranged; but, in this inquiry, we must particularly consider, whether the letters in question are the same, or different: for when they are the same, there can be no transposition of them; and for this reason the simple powers, as $$a^2$$, $$a^3$$, $$a^4$$, etc. have all unity for their coefficients.

355 Let us first suppose all the letters different; and, beginning with the simplest case of two letters, or $$ab$$, we immediately discover that two transpositions may take place, namely, $$ab$$ and $$ba$$.

If we have three letters, $$abc$$, to consider, we observe that each of the three may take the first place, while the two others will admit of two transpositions; thus, if $$a$$ be the first letter, we have two arrangements $$abc$$, $$acb$$; if $$b$$ be in the first place, we have the arrangements $$bac$$, $$bca$$; lastly, if $$c$$ occupy the first place, we have also two arrangements, namely, $$cab$$, $$cba$$; consequently the whole number of arrangements is 3 · 2 = 6.

If there be four letters $$abcd$$, each may occupy the first place; and in every case the three others may form six different arrangements, as we have just seen; therefore the whole number of transpositions is 4 · 6 = 24 = 4 · 3 · 2 · 1.

If we have five letters, $$abcde$$, each of the five may be the first, and the four others will admit of twenty-four transpositions; so that the whole number of transpositions will be 5 · 24 = 120 = 5 · 4 · 3 · 2 · 1.

356 Consequently, however great the number of letters may be, it is evident, provided they are all different, that we may easily determine the number of transpositions, and that we may for this purpose make use of the following Table:

Number of Letters. Number of Transpositions.
1 1 =1.
2 2 · 1 =2.
3 3 · 2 · 1 =6.
4 4 · 3 · 2 · 1 =24.
5 5 · 4 · 3 · 2 · 1 =120.
6 6 · 5 · 4 · 3 · 2 · 1 =720.
7 7 · 6 · 5 · 4 · 3 · 2 · 1 =5040.
8 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 =40320.
9 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 =362880.
10 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 =3628800.

357 But, as we have intimated, the numbers in this Table can be made use of only when all the letters are different; for if two or more of them are alike, the number of transpositions becomes much less; and if all the letters are the same, we have only one arrangement: we have therefore now shown how the numbers in the Table are to be diminished, according to the number of letters that are alike.

358 When two letters are given, and those letters are the same, the two arrangements are reduced to one, and consequently the number, which we have found above, is reduced to the half; that is to say, it must be divided by 2. If we have three letters alike, the six transpositions are reduced to one; whence it follows that the numbers in the Table must be divided by 6 = 3 · 2 · 1; and, for the same reason, if four letters are alike, we must divide the numbers found by 24, or 4 · 3 · 2 · 1, etc.

It is easy therefore to find how many transpositions the letters aaabbc, for example, may undergo. They are in number 6, and consequently, if they were all different, they would admit of 6 · 5 · 4 · 3 · 2 · 1 transpositions; but since a is found thrice in those letters, we must divide that number of transpositions by 3 · 2 · 1; and since b occurs twice, we must again divide it by 2 · 1: the number of transpositions required will therefore be

$\dfrac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1 \cdot 2 \cdot 1} = 5 \cdot 4 \cdot 3 = 60.$

359 We may now readily determine the coefficients of all the terms of any power; as for example of the seventh power $$(a + b)^7$$.

The first term is $$a^7$$, which occurs only once; and as all the other terms have each seven letters, it follows that the number of transpositions for each term would be

7 · 6 · 5 · 4 · 3 · 2 · 1,

if all the letters were different; but since in the second term, $$a^6b$$, we find six letters alike, we must divide the above product by 6 · 5 · 4 · 3 · 2 · 1, whence it follows that the coefficient is

$\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4 \cdot 3\cdot 2\cdot 1}=\frac{7}{1}.$

In the third term, $$a^5b^2$$, we find the same letter $$a$$ five times, and the same letter $$b$$ twice; we must therefore divide that number first by 5 · 4 · 3 · 2 · 1, and then by 2 · 1, whence results the coefficient

$\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4 \cdot 3\cdot 2\cdot 1 \cdot 2\cdot 1}=\dfrac{7 \cdot 6}{1 \cdot 2}.$

The fourth term $$a^4b^3$$ contains the letter $$a$$ four times, and the letter $$b$$ thrice; consequently, the whole number of the transpositions of the seven letters, must be divided, in the first place, by 4 · 3 · 2 · 1, and, secondly, by 3 · 2 · 1, and the coefficient becomes

$=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{4 \cdot 3\cdot 2\cdot 1 \cdot 2\cdot 1}=\dfrac{7 \cdot 6\cdot 5}{1 \cdot 2 \cdot 3}.$

In the same manner, we find $$\dfrac{7\cdot 6\cdot 5\cdot 4}{1 \cdot 2\cdot 3\cdot 4}$$ for the coefficient of the fifth term, and so of the rest; by which the rule before given is demonstrated.

360 These considerations carry us farther, and show us also how to find all the powers of roots composed of more than two terms. We shall apply them to the third power of $$a + b + c$$; the terms of which must be formed by all the possible combinations of three letters, each term having for its coefficient the number of its transpositions, as shown, Art. 352.

Here, without performing the multiplication, the third power of $$(a + b+ c)$$ will be,

$a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3.$

Suppose $$a=1$$, $$b=1$$, $$c=1$$, the cube of 1 + 1 + 1, or 3, will be

1 + 3 + 3 + 3 + 6 + 3 + 1 + 3 + 3 + 1 = 27;

which result is accurate, and confirms the rule. But if we had supposed $$a=1$$, $$b=1$$, and $$c=-1$$, we should have found for the cube of 1 + 1 - 1, that is of 1,

1 + 3 - 3 + 3 - 6 + 3 + 1 - 3 + 3 - 1,

which is a still farther confirmation of the rule.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.