Homework #8 — Math 173 — Spring ’09
Solutions
(1) 6.3; 4) a) Since there is only one parameter we are keeping track of (namely, r), we can
do this all with one variable (we’ll use zto avoid confusion with the equation variables).
Think of choosing balls from three boxes. In the first box are balls weighing two pounds
each. In the second and third boxes the balls weigh three and seven, respectively. We want
to know how many ways we can get rpounds total.
We can pick 0,1,2, . . . balls from the first box. In terms of choosing weights, these choices
correspond to 0,2,4, . . . pounds from the first box. So we want the GF 1 + x2+x4+·· · =
1/(1 −x2). Arguing similarly for the other boxes, we end up with 1
(1−x2)(1−x3)(1−x7).
b) Take the choice from the first box. We need to choose at least eight balls; this
corresponds to choosing at least 16 pounds. We want (x2)8+ (x2)9+· · · for this box.
Arguing similarly for the other boxes, we get
((x2)8+ (x2)9+· · · )((x3)2+· · · + (x3)8)(1 + (x7)1+ (x7)2)
for our final answer.
(2) 6.3; 6a) If the boxes were distinct, then this would just be a balls and boxes problem. But
since they’re indistinguishable, we’re going to overcount. For instance, as a balls-and-boxes
problem 7,4 and 1 balls going in each box is going to be counted differently from putting
4,7,1. Note that we can’t just divide by 3! as some configurations (such as 4,4,4) are not
going to be overcounted.
The solution is to recognize this question as asking for the number of partitions of rwith
at most three parts. By the transposition of Ferrers diagrams, this is equal to the number
of partitions of rwith parts of size at most 3. But from the derivation of the GF for all
partitions, we can easily construct this as 1
(1−x)(1−x2)(1−x3).
(3) 6.3; 11) a) Consider an arbitrary partition λ= (λ1, . . . , λk) (with λ1≥λ2≥ · · · ≥ λk>0).
If λk= 1, then (λ1, . . . , λk−1) is a partition of r−1 into k−1 parts. There are R(r−1, k −1)
such partitions. If λk>1, then (λ1−1, λ2−1, . . . , λk−1) is a partition of r−kinto kparts
(because λk−1≥1). There are R(r−k, k) such partitions. Since each of these processes
is reversible, we get the equation claimed.
Here is an example for n= 7 and k= 3. We have R(7,3) = 4 as the possible partitions are
, ,,. In removing the last box from each of the first three partitions,
we get the three possible partitions of 6 into two parts: , , and . For ,
we remove one from each part to get , the single partition of 4 into three parts.
b) Let λ= (λ1, . . . , λr) be a partition of ninto rparts with λk>1 and λk+1, λk+2, . . . , λr=
1. Remove 1 from each part. We now have a new partition (λ1−1,λ2−1, . . . , λk−1) of
n−rinto kparts. Conversely, if we have a partition of n−rinto kparts, adding r1’s
reverses the above property. The summation equality claimed thereby holds.
(4) 6.3; 18) As per the hint, consider a row of n1’s. Between each pair of successive 1’s, we
can either place a + sign or a ,. By the MR, this leads to 2n−1possible ordered partitions.
(5) 6.3; 21a) The book answers this pretty well, but here’s another description. Strip off the
northwest boundary of the Ferrers diagram and make this the first row of a new partition.
Since the original diagram was self-conjugate, this boundary has odd length (it consists of
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