1 Answers to Chapter 8, Odd-numbered Exercises

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1 Answers to Chapter 8, Odd-numbered Exercises ) (a) + 4x + 6x + 4x 3 + x 4. (b) + x + x + x 3 + x 4 + x 7. (c) x 3 + x 4 + 3x 5 + 4x 6 + 6x 7. (d) + x + x + x 3 + x 4 + x 5 + x 6. (e) 3 + x 3 4x 4 + 7x 5. (f) x 4 + x 5 3x 6 + x 8. 3) (a) By the Binomial Theorem, for 0 n, a n ( 0 n ). For n >, an 0. (c) For n 4k + 3, k 0, c n. Otherwise, c n 0. The generating function looks like x 3 ( + x 4 + x 8 + x ) x 3 + x 7 + x + x 5 + x 4k+3 (e) e n for n 0 except for n 3, 4; e 3 e 4. We combine the following three generating functions: to get that + x x 4 x (g) g n ( 4) n for n 0. To see this, x x k x x x k k x 4 x x k k4 + x + x + x 3 + x 4 + x 5 + x k. u u n. Now substitute u 4x into the right hand side and left hand side. (i)i n for n 7k or n 7k + or n 7k + where k 0. We combine the following three generating functions: x 7 x x 7 x 7k x 7k+ x x x 7k+ 7

2 to get that x + x + x 7k + x 7k+ + x 7k+. x 7 5) There are 4 ways to get 0 balloons, and the generating function is x +x 3 +x 4. Since we want ( x) at least one white balloon and at least one gold balloon, we model each of these with the generating function x x x + x + x 3. Since we want at most two blue balloons, we model this with + x + x. Hence, the number of ways to create a bunch of n balloons is the coefficient of in the expansion of x x x x ( + x + x ) x + x 3 + x 4. ( x) To find this coefficient, observe that Hence, ( x) d dx x (n + ). x + x 3 + x 4 (n + )+ + (n + )+3 + (n + )+4 ( x) (n ) + (n ) + (n 3) x + 3x 3 + n3 (3n 6) n4 Hence, there are 3n 6 ways to get a bunch of n balloons. 7) The number of solutions is ( ) n+ 3. The number of solutions to x + x + x 3 + x 4 n is equal to the number of solutions to x + x + x 3 + x 4 + x 5 n. For x, x, x 3, x 4, x 5, the associated generating functions are, respectively, x ; x x + x x 4n ; 4 + x + x + x 3 x4 x ; x. n4 ;

3 The coefficient of in the product gives us the number of solutions: Now, since we have that x x x x ( x) x 4 6 d 3 dx 3 x x 6 x 4 x4 x d 3 dx 3 x 6 ( x) 4 x x ( x) 4 n(n )(n ) 3 n3 ( ) n ) ( + x) p. For each of the p students, a student can either be chosen or not. Hence, their associated generating function is + x. Therefore, the generating function for the number of ways to choose n students from p students is ( + x) p p ( ) p. n ) 03. For dollar coins, dollar bills, and $ bills, the associated generating functions are, respectively: x x x x n So, the number of ways to make change for $00 is the coefficient of x 00 in the expansion of By a partial fraction expansion, ( x) ( x ). ( x) ( x ) /4 x + /4 + x + / ( x) Hence, the coefficient of x 00 is + ( ) ( ) n + (n + ). 3) (a) b n x ( x 7 )( + 4x + 0x ). ( x) ( x )

4 The generating functions for chocolate bites, peanut butter cups, and peppermint candies are, respectively, + x + x + x 3 + x 4 + x 5 + x 6 x7 x x x x For fruit chews, the generating function is a bit more complicated. If there are no fruit chews in the bag, there is exactly way to do this. If there is fruit chew in the bag, since there are ( 4 flavors, there are 4 ways to do this. If there are fruit chews in the bag, there are 4 ) ( ways to do this (there are 4 ) ways if the flavors are different, and 4 ways if the flavors are the same). So the generating function for fruit chews is + 4x + 0x. The product of these is the generating function for b n. (b) We are looking for the smallest n such that b n 400. We use a computer algebra system (Wolfram Alpha) to find the partial fraction decomposition: x ( x 7 )( + 4x + 0x ) ( x) ( x ) So for n > 7, the coefficient of is x n x + 6x + 8x x x 5 + 4x 6 + 0x 7 8/4 x + 7/4 + x + 05/ ( x) ( ) n (n + ). Plugging this into a calculator, we see that this is more than 400 for n 3. So, 3 pieces of candy are needed. 5) (a) 3. The generating functions for pennies, nickels, dimes, and quarters are, respectively, The coefficient of x 95 in the expansion of x x n 0 x 5n 5 x x 0n 0 x x 5n. 5 ( x)( x 5 )( x 0 )( x 5 )

5 is the number of ways to make change for $0.95. Using a computer algebra system, we find that this is 3. (b) The generating functions for pennies, nickels, dimes, and quarters are, respectively, + x 5 + x0! + x 0 + x0! + x 5 + x50! + x + x! + xn + x5 + x30 + x75 + x5n + x0n + x5n One can then use a computer algebra system, cutting off any terms after x 95 to get the result. 7) A partition of 0 into odd parts consists of s, 3s, 5s, 7s, and 9s. The generating function for each of these is, respectively, + x + x + x 3 + x 0 + x 3 + x 6 + x 9 + x 5 + x 0 + x 7 + x 9. Hence, the coefficient of x 0 in the product of these gives the answer. Multiplying these polynomials together with a computer algebra system yields 0 as the coefficient of x 0. 9) n x n ) (a) a n 7 n, because (b) b n n(n )3 n, because e 7x + 7x + (7x)! + (7x)3 x e 3x (3x) n+ (3x) n (n )! n(n )(3x) n (c) c n. 3) (4n 3 n n + ). The exponential generating functions for the letters a, b, c, d are, respec-

6 tively, e x e x e x e x e x n x k+ (k + )! The coefficient of / in the product gives the number of strings: (e x )(e x )( ex e x )(e x ) (e 3x e x )( ex e x ) (e4x e 3x e x + e x ) (4 n 3 n n + ). 5) a 0 0, a 0, a 0 and for n 3, a n n n n + ( ) n n (n)(n ) + ( ) n n(n ) The exponential generating functions for the letters a, b, c, d are, respectively, e x e x e x e x x + x. n x k+ (k + )! The coefficient of / in the product gives the number of strings: (e x )( ex e x )(e x )(x + x ) 4 (ex e x + e x )(x + x ) xe x x xe x + xe x + x e x x x e x + x e x x x + + n n ( ) n (n )! n (n )! (n )! n + (n )! + ( ) n (n )!