20. In how many terms in the preceding example, will a3 stand first? SUGGESTION.-The number will be equal to the permutations taken all together, of the letters in b4c2. Ans. 15. 21. In the permutations formed out of a, b, c, d, e, f, g, taken all together, how many begin with ab? How many with abc? How many with abcd? Ans. (1)120. (2)24. (3)6. 22. Out of 17 consonants and 5 vowels, how many words can be formed, having two consonants and one vowel in each? Ans. 4080. 23. Find the number of combinations that can be formed out of the letters of the word "Notation," taken 3 together. Ans. 22. BINOMIAL THEOREM, WHEN THE EXPONENT IS A POSITIVE INTEGER. ART. 310. We have already explained (Art. 172) the method of finding any power of a binomial, by repeated multiplication; and we shall now proceed from the theory of Combinations (Art. 308), to derive a general rule, which is called the Binomial Theorem, and sometimes Sir Isaac Newton's Theorem, from the name of the inventor. In its most general form the Binomial Theorem teaches the method of developing into a series any binomial whose index is either integral or fractional, positive or negative; that is, quantities of the form (a+x)”, (a+x) ̄”, (a+x)TM, (a+x)-” where a or x may be either plus or minus. The following investigation applies only to the case where the exponent is positive and integral, the other cases will be considered hereafter. (See Art. 319.) By actual multiplication it appears that Also, (x+a)(x+b)(x+c)(x+d) = x2+ax3+ ab x2+abc|x+abcd. +bc +bcd An examination of either of these products, shows that it is composed of a series of descending powers of x, and of certain coëfficients, formed according to the following law: 1st. The exponent of the highest power of x is the same as the number of binomial factors, and the other exponents of a decrease by 1 in each succeeding term. 2nd. The coëfficient of the first term is 1; of the second, the sum of the quantities a, b, c, &c.; of the third, the sum of the products of every two of the quantities a, b, c, &c. ; of the fourth, the sum of the products of every three, and so on; and of the last, the product of all the n quantities a, b, c, &c. Suppose, then, this law to hold for the product of n binomial factors x+a, x+b, x+c, x+k; so that (x+a)(x+b) (x+c).....(x+k)=x+Ax-1+Bx-2+Cx2-3+. . . . +K, If we multiply both sides of this equation by a new factor x+1, we have (x+a)(x+b)(x+c)..... (x+k)(x+1) ++A+B+C2-2 ... +Kl. It is evident the same law still holds; that is, 1st. The exponent of the highest power of x is the same as the number of binomial factors; and the other exponents of a decrease by 1 in each succeeding term, 2nd. The coefficient of the first term is 1. A+1, the coefficient of the second term, is the sum of the second terms, a, b, c, k, and l of the binomial factors. B+Al, the coefficient of the third term, is the sum of the products of the second terms of the binomial factors taken two together; for by hypothesis, B is the sum of the products of n binomial factors, taken two together, and Al is the product of the second terms of the preceding n binomials by the second term 7 of the new binomial; therefore, B+Al is the product of the second terms of all the binomial factors taken two together. Kl, the last term, is the product of all the second terms of the n+1 binomial factors. Hence, if the law holds when n binomial factors are multiplied together, it will hold when n+1 factors are multiplied together; but it has been shown by actual multiplication to hold up to 4 factors; therefore it is true for 4+1, that is 5; and if for 5, then also for 5+1, that is 6; and so on generally, for any number whatever. Now let then b, c, d, &c., each =a, A=a+a+a+a+, &c., to n terms =na. B=a2+a2+&c., a2 taken as many times as is equal to the No. of combinations of n things taken two together, which is (Art. 308), C=a3+a3+, &c., a3 taken as many times as is equal to the No. of combinations of the things taken three together, which is (Art. 308), _n(n-1)a2 1.2 _n(n-1)(n-2)as 1.2.3 (x+a)(x+a)(x+a) (x+1) becomes 1.2.3 +an. .. (x+a)"=x+nax2-1+ n(n−1) a2xn−2+n(n−1)(n—2) a3.xn-s 1.2 Let a 1, then since every power of 1 is 1, Cor. 1.- It is obvious that the sum of the exponents of a and x in each term =n. Cor. 2.-If either term of the binomial is negative, every odd power of that term will be negative (Art. 193); therefore the signs of the terms in which the odd powers are found will be negative. Here it is evident the coëfficient of any term is formed of the n product of the factors 2,21, 22, &c., in number one less 3 than the number which denotes the place of the term; therefore, the coefficient of the rth term will be Also, the exponent of x is the same as the denominator of the last factor of the coëfficient; and the exponent of a is equal to n minus the exponent of x, (Cor. 1); therefore, the whole rth term is, n(n-1)(n-2) (n—r+2) an-r+12-1. as above (-1) This is called the general term, because by making r=2, 3, 4, &c., all the others can be deduced from it. Ex. Required the 5th term of (a—x). Cor. 4.- If n be a positive integer, and r=n+2, then (n-r+2) becomes 0, and the (n+2) term vanishes; therefore, the series consists of (n+1) terms altogether; that is, in raising a binomial to any given power, the number of terms is one greater than the exponent of the power to which the binomial is to be raised. Cor. 5.-When the index of the binomial is a positive integer, the coefficients of the terms taken in an inverse order from the end of the series, are equal to the coefficients of the corresponding terms taken in a direct order from the beginning. If we compare the expansion of (a+x)", and (x+a)", we have (a+x)"=a"+na"-1x+n(n−1), c2+ _n(n−1)(n—2) an Lan-3x2+, &c 3+,&c 1.2 1.2.3 Since the binomials are the same, the series resulting from their expansion must be the same, except that the order of the terms will be inverted. It is clearly seen that the coëfficients of the corresponding terms are equal. Hence, in expanding a binomial, whose index is a positive integer, the latter half of the expansion may be taken from the first half. Ex. Expand (a—b)5. Here the number of terms (n+1) is equal to 6; therefore, it will only be necessary to calculate the coefficients of the first three, thus: Cor. 6.-The sum of the coefficients of any expanded binomial whose index is n, and where both terms are positive, is always equal to 2". For if x-a-1, then (x+a)" (1+1)"=2" _n___n(n−1)___n(n−1)(n—2)+n(n−1)(n—2)(n—3)+, &c. =1+2+77-12+ 1.2 1 2 3 1.2.3.4 |