
Chapter 4
 #16: In a standard deck of 52 cards,
there are 13 cards of each of four suits. To
collect a set of four cards each of different
suits, we must select one card from each suit.
This can be done in 13×13×13×13=13^{4}
ways.
Chapter 5 Section 1
 #5: (a) and (b): Because "no family was
completely without such items," we know all 983
families had at least one of the items. We now
compare information about the number of families
with known numbers of items. Because 345 families
had at least 3 items, and 264 families had all 4
items, the difference, 81 families tells us the
number of families with exactly 3 items. Using
this same process, we can determine that 136
families possessed exactly 2 items, and 502
families possessed exactly 1 item. Note that
502+136+81+264 = 983. Use this four values to
determine the total number of items held by all
983 families:
 502×1 + 136×2 + 81×3 + 264×4 = 2073 items
Chapter 5 Section 2
 #1: This is the number of derangements of n
items: D(n) = n!(1  1/1! + 1/2!  ... + (1)^{n}/n!)
 #2: D(6) = 265
Chapter 6
 #2: s(0)=5, s(1)=2(5)+3=13, s(2)=32, s(3)=73,
s(4)=158, s(5)=331
 #5: s(0)=1, s(1)=2, s(2)=5, s(3)=5+2(2)+1+3=13,
s(4)=29, s(5)=65
 #8: The word "sequence" implies that order
matters.
 $1: either Tape (T) or Ruler (R), so S(1)=2
 $2: TT, RR, TR, RT, or Pen (P1), Pencil (P2),
Paper (P3), so S(2)=7. Note that this is
2·S(1)+3.
 $3: To each $1 purchase, we can attach one of
three $2 items; to each $2 purchase, we can one
of two $1 items; we can also purchase a Binder
(B). So S(3) = 21. Note that this is
3·S(1)+2·S(2)+1.
 $4: 2·S(3)+3·S(2)+S(1)=65
 So for $N, we have the recurrence relation
S(N) = 2·S(N1) + 3·S(N2) + S(N3), with
initial conditions that S(0) = 1, S(1) = 2, and
S(2) = 7.
 #11: From previous activities, we know that
there are 2^{n} subsets for a set with n
elements. So for n = 1,2,3,..., we have s(n) =
2,4,8,.... Recursively, this means s(n) = 2·s(n1)
with s(1) = 2.
 #15:The word "select" indicates that the order
items are chosen is not significant. Given the
kind and number of fruits available, the fruit
pairs include AA, AP, AO, AB, PO, PB, OB. This is
a collection 7 fruit pairs.
 #16: We assume here
that each person wants one fruit danish, and
that we are not concerned about which person
ordered which danish, merely that orders were
placed.
 0 people ordering: 1 way. d(0) = 1
 1 person ordering: 3 choices, so there are
three possible orders. d(1) = 3
 2 people: Because there are at least two of
every danish, we can consider all ways to
collect two danish: CC,AA,RR,CA,CR,AR. d(2) =
6.
 3 people: Now we must be concerned not to
sell more than 2 apricot danish among the
possible orders:
CCC,RRR,CCA,CCR,AAC,AAR,RRC,RRA,CAR. d(3) = 9.
 4 people: Now we must be concerned not to
sell more than 3 cheese danish as well as only
2 apricot danish among the possible orders:
RRRR,CCCR,CCCA,RRRC,RRRA,RRCC,RRAA,CCAA,RRCA,CCAR,AACR.
d(4) = 11.
 Now we begin to use the symmetry for set
inclusion/exclusion. For instance, because
only 9 danish are available, The 11 orders
possible for 4 people, d(4) = 11, means 5
danish are not ordered. Thus, d(5) = 11, d(6)
= 9, d(7) = 6, d(8) = 3, and d(9) = 1.
 #17: In determining whether order is
significant, consider the situation. If I spend
$7, I can get two cheese sandwiches and one bowl
of soup. Does the order one arranges those items
(CCS, CSC, SCC) make this three distinct orders?
No, the order is two cheese sandwiches and one
bowl of soup. So, here, the arrangement of the
items within a food order is not significant. By
brute force enumeration, we have:
 d(0) = d(1) = 0: There is no way to place an
order whose value is $0 or $1.
 d(2) = 1
 d(3) = 1
 d(4) = 1
 d(5) = 1
 d(6) = 2
 d(7) = 1
 d(8) = 2
 d(9) = 2
 d(10) = 2
 d(11) = 2
 d(12) = 3
 #20: For the sequence 1,2,4,8,16, . . . , each
new term is twice the previous: s(n) = 2·s(n1),
with s(1) = 1
 #32: We seek a recursive relationship
so that the number of rectangles
measuring 3byn can be described in terms of
the number of rectangles of smaller length. We
observe that there is one way to create a 3by1
arrangement, two ways to create a 3by2
arrangement and six ways to create a 3by3
arrangement:
 Now we can create new 3byn
arrangements using these. For the 3by4
arrangements, first use the 3by1
arrangement and attach to it each of the
last (bottom) three 3by3 arrangements
shown above. Next, use the each existing
3by2 arrangement and attach a 3by2
tile. Finally, to each of the six 3by3
arrangments, attach a 3by1 tile. So,
A(4) = 3·A(1) + A(2) + A(3).
 This recursion can be generalized to A(n)
= 3·A(n3) + A(n2) + A(n1), with A(1)
= 1, A(2) = 2, and A(3) = 6.

