Suggested Solutions — Math 173 — February 26, 2009
(1) 6.1; 25) Let’s focus on one child. We can offer any number of chocolate bars (yielding a mini-
GF of (1+x+x2+· · · )) and at most three lollipops (yielding a mini-GF of (1+y+y2+y3)).
Since these are independent choices, for each child we have the product of these two as our
GF. Since there are five children, we just raise this product to the fifth power to get a final
answer of
(1 + x+x2+· · · )(1 + y+y2+y3)5=1−y4
(1 −x)(1 −y)5
.
(2) 8.2; 13) Let Aibe the set of choices consisting of exactly two pieces of the candy of type
i. Suppose we know the child takes exactly two pieces of ktypes of candy (and perhaps of
other types we don’t know about). There are 4
kways to choose these types. The remaining
ways to pick the other 11 −2kpieces is just a balls and boxes problem with r= 11 −2k
and n= 4 −k. Plugging into oru inclusion-exclusion formula, we get
11 + 4 −1
11 −4
19+3−1
9+4
27+2−1
7−4
35+1−1
5.
Note that the answer in the book mistakenly leaves off the last term.
(3) 8.2; 19) What could go wrong is that a professor could teach the same pair of courses both
semesters. So let Aibe the set of assignments for which the i-th professor teaches the same
pair both semesters. We wish to count
(1) |¯
A1∩ · · · ¯
A5|=|U| − S1+S2−S3+S4−S5.
|U|can be computed as a standard “permutation with restricted repetition problem.” Write
down all ten courses, say ABCDEFGHIJ. Then write down a word consisting of two 1’s,
two 2’s, .. . , two 5’s; say 5311423542. We can interpret this word as assigning courses C&D
to professor 1, courses F&J to professor 2, etc. for a given semester. The number of such
words is of course P(10; 2,2,2,2,2) = 10!/25. Since there are two semesters, we find that
|U|= (10!)2/210.
Now, suppose that we are in a k-intersection (i.e., we are aware of kprofessors getting
the same assignments both semesters; there may be others as well!). We pick the fall
assignments as above in one of (10!)/25possible ways. Then we pick which kprofessors
have a repeated pair in 5
kways. We don’t have any choice for the spring assignments for
these professors. But we do make the spring assignments for the 5 −kother professors in
one of (10 −2k)!/25−kpossible ways. Plugging into (1), we get
5
X
k=0 5
k(−1)k10!
25
(10 −2k)!
25−k.
(4) 8.2; 37) Both parts are easily solved by Theorem 2. I won’t be asking you such a problem
on the exam(s). I just want you to know the existence of these variations. It’s still a useful
exercise to solve this problem.
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