AMS 301 - Homework 6 Solutions

6.1.2

a. We want 5 product terms, each of which has all the powers of x between 0 and 4, inclusive.Therefore, g(x) = (1 + x + x2 + x3 + x4)5

b. We want 3 product terms, each of which has all the powers of x between 0 and 5, exclusive.Therefore, g(x) = (x + x2 + x3 + x4)3

c. We want 4 product terms, each with only powers between 2 and 6, inclusive, but with theextra constraint that the first uses only even powers and the second uses only odd powers.Therefore, g(x) = (x2 + x4 + x6)(x3 + x5)(x2 + x3 + x4 + x5 + x6)2.

d. We want 4 product terms, each using all non-negative powers. Therefore, g(x) = (1 + x +x2 + x3 + . . . )4

e. We want 4 product terms, each using only positive powers, with the extra constraints thatthe second and fourth only use odd powers and the fourth only uses powers less than or equalto 3. Therefore, g(x) = (x + x2 + x3 + x4 + . . . )2(x + x3 + x5 + x7 + . . . )(x + x3)

6.1.4

a. g(x) = (1 + x + x2 + x3 + x4 + x5)5

b. g(x) = (x3 + x4 + x5 + x6)4

c. g(x) = (x + x2 + x3 + x4 + . . . )7

d. g(x) = (1 + x + x2 + x3 + x4 + x5)(1 + x + x2 + x3 + . . . )2

6.1.16

We want a number with digits e1 + e2 + e3 + e4 + e5 + e6 = r. The constraints will be 0 ≤ ei ≤ 9.Therefore, the generating function is g(x) = (1 + x + x2 + · · ·+ x8 + x9)6.

6.2.2

We want the coefficient of xr in g(x) = (x5 + x6 + . . . )7. First, factor out the power of x inthe first term. Don’t forget to raise it to the correct power! g(x) = (x5)7(1 + x + x2 + . . . )7 =x35(1 +x+x2 + . . . )7. Then use the formula for infinite geometric series, and say g(x) = x35 1

(1−x)7.

Use the formula for 1(1−x)m to get g(x) = x35(1+C(7, 1)x+C(8, 2)x2 + · · ·+C(k + 6, k)xk + . . . ). To

find the coefficient of xr, we want to find the term in the second product that has k = r − 35, sothe coefficient would be C(r − 35 + 6, r − 35) =C(r − 29, r − 35).

6.2.6

We want the coefficient of x52 in g(x) = (x10 +x11 + · · ·+x25)(x+x2 + . . . x15)(x20 +x21 + · · ·+x45).First factor out the power of x in the first term of each product. g(x) = x31(1 + x + . . . x15)(1 +x + · · · + x14)(1 + x + · · · + x25). Then use the finite geometric series formula, to get g(x) =x31(1−x16

1−x )(1−x15

1−x )(1−x26

1−x ) = x31(1−x16)(1−x15)(1−x25)( 1(1−x)3

). Then use the formula for 1(1−x)m

to get g(x) = x31(1 − x16)(1 − x15)(1 − x25)(1+C(3, 1)x+C(4, 2)x2 + · · ·+C(r + 2, r)xr + . . . ).

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We are looking for the coefficient of x52 in g(x), so note that the first product introduces x31,and the next 3 terms all introduce either x0, or xk. The third term cannot introducte the x25 asx31+25 = x56 which is already a power greater than 52. We also cannot choose both the x15 andthe x16 as x31+15+16 = x62. So there are 3 terms of importance for the coefficient of x52. They arex31(1)(1)(1)(C(23, 21)x21), x31(−x16)(1)(C(7, 5)x5), and x31(1)(−x15)(C(8, 6)x6). Therefore, thecoefficient of x52 is C(23, 21)−C(7, 5)−C(8, 6).

6.2.22

We want to find the number of ways to get a sum of 25 when 10 dice are rolled. This is equivalentto asking for the number of integer solutions to e1 + e2 + · · · + e10 = 25 for 1 ≤ ei ≤ 6. Thishas a generating function of g(x) = (x + x2 + x3 + x4 + x5 + x6)10. Factor out the x from eachterm to get g(x) = x10(1 + x + x2 + x3 + x4 + x5)10. Use the finite geometric series formula to getg(x) = x10(1−x6

1−x )10 = x10(1− x6)10 1(1−x)10

. Then use the formulas for (1− xm)n and 1(1−x)m to get

g(x) = x10(1−C(10, 1)x6+C(10, 2)x12−C(10, 3)x18 + · · ·+C(10, 10)x60)(1+C(10, 1)x+C(11, 2)x2 +· · ·+C(r + 9, r)xr + . . . ). We are looking for the coefficient of x25, so the important terms arex10(1)(C(15 + 9, 15)x15), x10(−C(10, 1)x6)(C(9 + 9, 9)x9, and x10(C(10, 2)x12)(C(3 + 9, 3)x3). Thismeans that the coefficient of x25 is C(24, 15)−C(10, 1)C(18, 9)+C(10, 2)C(12, 3), which is the num-ber of ways of getting a sum of 25 after 10 dice are rolled.

6.3.1

a. There are 5 partitions of 4 and they are 1 + 1 + 1 + 1, 2 + 1 + 1, 2 + 2, 3 + 1, and 4.

b. There are 11 partitions of 6 and they are 1 + 1 + 1 + 1 + 1 + 1, 2 + 1 + 1 + 1 + 1, 2 + 2 + 1 + 1,2 + 2 + 2, 3 + 1 + 1 + 1, 3 + 2 + 1, 3 + 3, 4 + 1 + 1, 4 + 2, 5 + 1, 6

6.3.15

To show this, we show that every partition of n corresponds to a unique partition of 2n into n parts,and vice versa. Take a partition of n, say a1+a2+· · ·+ak. We must have that k ≤ n. We correspondthis partition of n to the partition of 2n that is (a1 + 1) + (a2 + 1) + . . . (ak + 1) + 1 + 1 + 1 · · ·+ 1,where there are n−k ones. Note that this is indeed a partition of 2n, we added exactly n ones to apartition of n. It also has exactly n parts, k from the original partition and n− k new ones. We’vecorresponded a partition of n to a partition of 2n into n parts. The correspondence is unique, aswe can revert back to the original partition of n by subtracting 1 from every term and discardingthe zeros. Therefore, for every partition of n there is exactly one partition of 2n into n parts, andvice versa. This means that there are the same number of partitions of n as there are partitions of2n into n parts.

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