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To simplify the expression chronic gastritis message boards order ranitidine with a visa, we extract x 2 from each polynomial factor and then apply identity (2) gastritis diet ��������� best buy ranitidine. But the coefficient of x 16 in this latter expression will be the coefficient of x 6 in (1 - x)-5 [i gastritis aguda discount 150 mg ranitidine overnight delivery. From expansion (5) gastritis green tea buy 150 mg ranitidine otc, we see that the coefficient of x 6 in (1 - x)-5 is C(6 + 5 - 1, 6). More generally, the coefficient of x r in x 10 (1 - x)-5 equals the coefficient of r -10 x in (1 - x)-5 -namely, C((r - 10) + 5 - 1, (r - 10)). Observe that (x 2 + x 3 + x 4 + · · ·)5 is the generating function ar, for the number of ways to select r objects with repetition from five types with at least two of each type. In the last chapter, we solved such a problem by first picking two objects in each type-one way-and then counting the ways to select the remaining r - 10 objects-C((r - 10) + 5 - 1, (r - 10)) ways. In the generating function analysis in Example 1, we algebraically picked out an x 2 from each factor for a total of x 10 and then found the coefficient of x r -10 in (1 + x + x 2 + · · ·)5, the generating function for selection with unrestricted repetition of r - 10 from five types. Such correspondences are a major reason for using generating functions: the algebraic techniques automatically do the combinatorial reasoning for us. Example 2 Use generating functions to find the number of ways to collect \$15 from 20 distinct people if each of the first 19 people can give a dollar (or nothing) and the twentieth person can give either \$1 or \$5 (or nothing). This collection problem is equivalent to finding the number of integer solutions to x1 + x2 + · · · + x19 + x20 = 20 when xi = 0 or 1, i = 1, 2. The generating function for this integer-solution-of-an-equation problem is (1 + x)19 (1 + x + x 5). The first part of this generating function has the binomial expansion (1 + x)19 = 1 + 19 19 2 19 r 19 19 x+ x +···+ x +···+ x 1 2 r 19 If we let f (x) be this first polynomial and let g(x) = 1 + x + x 5, then we can use (6) to calculate the coefficient of x 15 in h(x) = f (x)g(x). Then the coefficient of x 15 in h(x) = f (x)g(x) is, by (6), a15 b0 + a14 b1 + a13 b2 + · · · + a0 b15 which reduces to a15 b0 + a14 b1 + a10 b5 since b0, b1, b5 are the only nonzero coefficients in g(x). Substituting the values of the various as and bs in (8), we have 19 19 19 19 19 19 Ч1+ Ч1+ Ч1 = + + 15 14 10 15 14 10 the answer in Example 2 could be obtained directly by breaking the collection problem into three cases depending on how much the twentieth person gives: \$0 or \$1 or \$5. In each case, the subproblem is counting the ways to pick a subset of the other 19 people to obtain the rest of the \$15. The generating function approach automatically breaks the problem into three cases and solves each, doing all the combinatorial reasoning for us. Example 3 How many ways are there to distribute 25 identical balls into seven distinct boxes if the first box can have no more than 10 balls but any number can go into each of the other six boxes? Using expansion (5), we have g(x) = (1 - x)-7 = 1 + 1+7-1 2+7-1 2 x+ x 1 2 r +7-1 r +···+ x +··· r We want the coefficient of x 25 (25 balls distributed) in h(x) = f (x)g(x). As in Example 2, we need to consider only the terms in the product of the two polynomials 1 7 (1 - x 11) and 1-x that yield an x 25 term. So the coefficient of x 25 in f (x)g(x) is a0 b25 + a11 b14 = 1 Ч 25 + 7 - 1 14 + 7 - 1 + (-1) Ч 25 14 the combinatorial interpretation of the answer in Example 3 is that we count all the ways to distribute without restriction the 25 balls into the seven boxes, C(25 + 7 - 1, 25) ways, and then subtract the distributions that violate the first box constraint, that is, distributions with at least 11 balls in the first box, C((25 - 11) + 7 - 1, (25 - 11)) (first put 11 balls in the first box and then distribute the remaining balls arbitrarily). The next example employs all the techniques used in the first three examples to solve a problem that cannot be solved by the combinatorial methods of the previous chapter. Example 4 How many ways are there to select 25 toys from seven types of toys with between two and six of each type? The generating function for ar, the number of ways to select r toys from seven types with between two and six of each type, is (x 2 + x 3 + x 4 + x 5 + x 6)7 We want the coefficient of x 25. As in Example 1, we extract x 2 from each factor to get [x 2 (1 + x + x 2 + x 3 + x 4)]7 = x 14 (1 + x + x 2 + x 3 + x 4)7   