22. Find the number of combinations of (a+b+c) things taken r together, a of the things being of one sort, b of another, and call different. 23. There is an army consisting of 10 European and 20 Sepoy regiments, and it is required to divide it into three detachments of 8, 10, and 12 regiments respectively, and such that the number of Europeans in any detachment shall never be less than one-fourth of the whole number of men in the detachment: find in how many ways this may be done. CHAPTER XVI THE BINOMIAL THEOREM. 165. In such expressions as (1 + x)3, (a — b)20, (3ab 3 + 2 - &c., the index denotes that a certain operation is to be performed upon the Binomial within the brackets; when this operation is completed, the Binomial is said to be expanded. Thus 13x + 3x2 + x3 is called the expansion of (1 + x)3. The "Binomial Theorem" is a formula for finding the terms in the expansion of a Binomal, without multiplication or extraction of roots. It is one of the discoveries of Sir Isaac Newton. be taken, we shall have a result of the following form: It is evident :-That the first term is unity, for it results from the multiplication of n factors, each of which is unity. That the second line contains the n letters a, b, c, d, &c., for each term results from the multiplication of n factors, (n − 1) of which are unity, and the other is a single letter; there are therefore as many terms as there are different letters. That the third line contains all the combinations that can be made of the n letters a, b, c, d, &c., taken two together, for each term results from the multiplication of n factors, (n − 2) of which are unity, and the other two are some combination of the n letters taken two together. And generally, that where each term contains r letters, the number of terms in that line must be the number of combinations of ʼn things taken r together, or Also the last line contains one term, for it is the product of the n letters, and there is only one combination of ʼn things taken altogether. Now the number of terms in each line will be the same, if a, b, c, d, &c. be all equal to one another. Let a = b = c=d, &c. = x suppose. Each of the factors to be multiplied becomes 1 + x, and the product of the n factors is represented by (1 + x)", and the above lines of products become respectively Note. In formulæ (1) and (2) the following points should be observed and remembered : 1o. The number of terms in the expansion is one more than the index. 2o. The index of x (the second term of the Binomial) increases by one in every successive term of the expansion. 3o. The coefficients of the terms equidistant from the beginning and end are alike. 40. The last factor in the denominator of each coefficient is the same as the index of x in that term. 5°. In formula (2) the sum of the indices of a and x in any term is equal to n, the original index. Some of the most useful applications of this theorem will be seen in the examples worked out below. = a + 18a5x + 135a4x2 + 540a3x3 + 1215a2x4 +1458αx+729x6. Note.-In expanding (a2 - 3)12 it should be written in the form {a2 + (-b3)}12. 3. What is the coefficient of x in the expansion of (1 + x2)10? The term involving 28 is the 5th term, and the coefficient of 12.11.10.9 this is 4. Find the 7th term, also the (n+1)th term in the expansion of The (n+1)th term is obtained from the above expression for the 7th term by writing (n + 1) instead of r throughout. Hence Note. The student should note the form (-1)-1; it is a convenient way of expressing that the sign of the term is positive or negative, according as r is odd or even. 5. Find the coefficient of x3 in the expansion of (1 + x + x2)10. Now (1 + x + x2)10 = {1 + x(x + 1)}10 = 1+10x (x + 1) + 45x (x + 1)2 +120 x (x + 1)3+, &c. = 1 + 10x (x + 1) + 45x2 (x + 1)2 It is evident that all the powers of x which appear after the 4th term are higher than 23; we must now select the terms involving 23 which arise by multiplying out the 3rd and 4th terms, and add them together. These are 45 x 2 + 120, Or, Or, 90+ 120; 210, the coefficient of x3. 6. Find the number of terms in the expansion of (a+b+c)13. The successive powers of (b+c) are now to be expanded, and the terms of each expansion are to be multiplied into the corresponding power of a outside the bracket. It is evident the whole number of terms will be 7. Find the greatest coefficient in the expansion of (3a + 2x)30. The coefficient of the 7th term of this is and the factor which is introduced to form the coefficient of the r+1 th than 1, the coefficients from this point cease to increase, or begin to decrease. To find the term which has the greatest coefficient we therefore put 31 |