Homework #7 — Math 173 — Spring ’09
Solutions
Homework
(1) 8.2; 32) Let Aibe the set of sequences in which player idoesn’t get to play :(. Suppose we
are in a k-intersection of the Ai’s, i.e., we know about kparticular players who don’t get to
play. The first match can be chosen in 6−k
2ways. The second match in 6−k
2−1 ways, etc.
In summary, all four matches can be chosen in P(6−k
2,4) ways. Backing up, there are 6
k
ways to choose the players we know aren’t playing. And kcan be 0, 1 or 2. (If k≥3, then
there are at most three pairs that can be made from the 6 −kplayers. This means we can’t
even have a different match each week.) Plugging into the inclusion-exclusion formula, we
get
P(6
2,4) −6
1P(6−1
2,4) + 6
2P(6−2
2,4).
(2) 6.2; 25) Let our exponents encode how many books the first teacher gets. For the first
book, we can give the teacher either 2, 3 or 4 copies (no more since the other teacher needs
to get at least two copies. Similarly, we can give 2 to 5 copies of the second book and 2 to
9 copies of the third book. Rewriting as GF’s, we have
[x12](x2+x3+x4)(x2+x3+x4+x5)(x2+x3+· · · +x9)
= [x6](1 + x+x2)(1 + x+x2+x3)(1 + x+· · · +x7).
At this point, I would just multiply it out; I’m not sure what the book did for its solution.
You get 12.
(3) 6.2; 29) Focus on one person. We can give the person anywhere from 0 to 5 of each delectable
type of food. Since we want to keep track of steak and lobsters separately, we’ll encode this
as
(1 + x+x2+x3+x4+x5)(1 + y+y2+y3+y4+y5).
There are four people, so our final equation will be
[x10y15](1 + x+x2+x3+x4+x5)(1 + y+y2+y3+y4+y5)4
= [x10](1 −x6)4
(1 −x)4[y15](1 −y6)4
(1 −y)4.
This can be solved as:
10 + 4 −1
10 −4
14+4−1
415 + 4 −1
15 −4
19+4−1
9+4
23+4−1
3.
(4) 6.1; 26bd)
(a)
x3y3+x4(y3+y4) + x5(y3+y4+y5) + x6(y3+y4+y5+y6)n(z3+z4+z5+z6)n.
(b)
(y3+y4+y5+y6)n(x3y4+x3y5+x3y6+x4y3+x4y5+x4y6+
x5y3+x5y4+x5y6+x6y3+x6y4+x6y5)n.
1