Elements of Algebra

116 ELEMENTS SECT. II. 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 $a^2, a^3, a^4$, &c. 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 $3 \times 2 = 6$. 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 $4 \times 6 = 24 = 4 \times 3 \times 2 \times 1$. If we have five letters, $abcde$, each of the five may be the