Elements of Algebra

118 ELEMENTS SECT. II. positions required will therefore be $\frac{6.5.4.3.2.1}{3.2.1.2.1} = 5$. 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 $\frac{7.6.5.4.3.2.1}{6.5.4.3.2.1} = \frac{7}{1}$, or $7$. 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 $\frac{7.6.5.4.3.2.1}{5.4.3.2.1.2.1} = \frac{7.6}{1.2} = 21$. 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 $=\frac{7.6.5.4.3.2.1}{4.3.2.1.3.2.1} = \frac{7.6.5}{1.2.3}$. In the same manner, we find $\frac{7.6.5.4}{1.2.3.4}$ 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 shew us * From the Theory of Combinations, also, are frequently deduced the rules that have just been considered for determining the coefficients of terms of the power of a binomial; and this is perhaps attended with some advantage, as the whole is then reduced to a single formula. In order to perceive the difference between permutations and combinations, it may be observed, that in the former we inquire in how many different ways the letters, which compose a certain formula, may change places; whereas, in combinations it is only necessary to know how many times these letters may be taken or multiplied together, one by one, two by two, three by three, \ldots