
Chapter 2 Review
 #5: Using the multiplication principle, there
are 6∙3∙8 = 144 meal combos. By the Pigeonhole
Principle, there must be 144 + 1 = 145 meal orders
placed to assure that at least one meal combo is
ordered twice.
 #6: Using the addition principle, there are 5 +
7 + 4 = 16 different breakfast sandwiches.
Chapter 3 Section 1
 #14: With no restrictions whatsoever, there are
26∙26 twoletter arrangements. There are
restrictions, however. There are 26 twoletter
arrangements with duplicate letters. There are 25
adjacent pairs, from ab and bc to
xy and yz. We double 25 because the
order of adjacency isn't restricted, so ba,
cb, . . . , yx, zy also
cannot be used. So from 26∙26 we remove (26 +
2∙25): 26∙26  26 − 2∙25 = 26∙(26−1) − 2∙25 =
26(25)−2(25) = 24(25) = 600.
Chapter 3 Section 2
 #17: For the set of digits
{0,1,2,3,4,5,6,7,8,9}, each digit either is in a
subset or it is not. Thus, for each digit, there
are two options for subset inclusion. By the
multiplication principle, there are 2^{10}
different subsets. We must exclude the empty set
and we must exclude the subset that contains all
ten digits. So we have 2^{10} − 2 = 2∙(2^{9}1)
different allowable subsets.
Chapter 3 Section 3
 #3: (a) Position the 13 nondistinguishable
mathematics (M) books. Among these books, there
are 14 places to position a science (S) book so it
will not be adjacent to other S books. So from
those 14 spots we choose 7: C(14,7). (b) If the
books can be distinguished, there are 13! ways to
first arrange the M books. Then, there are P(14,7)
ways to arrange the S books among these M books so
that no two S books are adjacent. We crossmatch
these options, using the multiplication principle,
to get 13!∙P(14,7) unique shelving arrangements.
 #5: (a) The path is 19 blocks long. Imagine a
19spot arrangement within which we position 12
identical characters W and 7 identical characters
S. There are C(19,12) ways to select spots for the
12 identical W's. Once positioned, there is only
one way, C(7,7), to position the identical S's.
Thus, there are C(19,12) paths for Geovanna. (b)
This restriction requires Geovanna to walk S
first. From there, she can walk any 18block path
that goes 6 blocks S and 12 blocks W. The number
of such paths, justified using the same argument
as in (a), is C(18,12). (c) The only way Geovanna
can walk a 21block path and go from home to the
library is to either (i) include one additional S
block then account for it with a N block, or, (ii)
include one additional W block and account for it
with an E block. Option (i) results in C(21,8)∙C(13,12)∙C(1,1)
and option (ii) results in C(21,13)∙C(8,7)∙C(1,1).
Because these two options result in disjoint
paths, there are C(21,8)∙C(13,12)∙C(1,1) +
C(21,13)∙C(8,7)∙C(1,1) such paths in all.
 #10: This is similar to the question posed in
class about birthorder arrangements, by gender,
in a family. We have 10 flips in all, with 6
identical HEAD (H) results and 4 identical TAIL
(T) results. There are C(10,6) = C(10,4)
arrangements. If k of the flips are T, we
have C(10,k) = C(10,10k)
arrangements.
 #18: In the word PREFERRER, there is one P, four
letters R, three appearances of E, and one F,
making 9 letters in all, four of which are
distinct. This is an arrangement when repetition
is allowed. We have 9!/(1!4!2!1!) possible
arrangements. This is the same as C(9,4)∙C(5,3)∙C(2,1)∙C(1,1).
 #31: We first pair up the paperbacks, place
them, and then insert hardback books between the
pairs to assure nonadjacency of the pairs. To
complete the first task, we can think of grabbing
two paperback books at a time and arranging them.
This can be done in C(6,2)*P(2,2) * C(4,2)*P(2,2)
* C(2,2)*P(2,2) ways. Simplifying this
expressions, we find it is equivalent to P(6,6) =
6!. Thus, we could think of completing the first
task as simply arranging the six paperback books.
We can consider arrangement of the three hardback
books in the same way. There are P(3,3) = 3! ways
to line up those books, regardless of where they
go. Now we must place them. What are the
possibilities for these three hardbacks? Let us
use PB to represent two paperback books together
and H to represent one hardback. We could have:
HPBHPBHPB or PBHPBHPBH or PBHHPBHPB
or PBHPBHHPB. Without changing the order of
the set of three hardbacks, we have four ways to
position the H books and still meet pairadjacency
requirements of the PB books. Remember, again,
that P(3,3)=3! had previously accounted for all
linear arrangements of the three HB books, so this
last sequence of counts was just about choosing
spots to assure the PB pairs restrictions. All of
these counts are matchups, so we multiply: 6!∙3!∙4.
 #33: Here we need to look at disjoint cases
determined by the number of games in a series.
There are four such cases, whose count results we
will add at the end. For each case, it's
significant to note that the series must end with
a W. That is, for the winning team, that team's
last game must have been a win.
 Case I: 4game sweep. There is only one way
for this to be accomplished: WWWW. In the format
of the following cases, that is C(3,3), with a W
required in 4th position.
 Case II: a 5game series. We must place a W in
the fifth position. Now we arrange the remaining
three W's among the first four positions:
C(4,3). With that task complete, the single L is
already determined.
 Case III: a 6game series: Again, we need a W
in the last position (game six). Now we arrange
the remaining three W's among the first five
positions: C(5,3). With that task complete, the
two L's are already determined.
 Case IV: a 7game series: Again, we need a W
in the last position (game seven). Now we
arrange the remaining three W's among the first
six positions: C(6,3). With that task complete,
the three L's are already determined.
 Now we add the counts from the four disjoint
cases: C(3,3) + C(4,3) + C(5,3) + C(6,3).
Chapter 3 Section 4
 #2: C(18,9) = C(17,9) + C(17,8)
 #6: We know C(10,5) = C(9,5) + C(9,4) using
Pascal's Formula. We manipulate this to get
C(10,5)  C(9,5) = C(9,4). The desired single
combination expression is C(9,4).
 #10: In (2a + 5b)^{n} there are n + 1
collected terms.
 #16: In the expansion of (a + b + c)^{8},
we seek the number of different arrangements of
the eight characters aaaabbbc. There are
8!/(4!3!1!) such unique arrangements.
 #21: (a) In (2a + b)^{5} there are n + 1
collected terms, with n = 5, so there are 6
collected terms. (b) The expansion and collection
begins with C(5,5)*(2a)^{5}*b^{0}
+ C(5,4)*(2a)^{4}*b^{1} +
C(5,3)*(2a)^{3}*b^{2} = 1*32a^{5}b^{0}
+ 5*16a^{4}b^{1} + 10*8a^{3}b^{2}
= 32a^{5} + 80a^{4}b
+ 80a^{3}b^{2}
 #32: In the expansion of (a + b + c + d)^{8},
we seek the number of different arrangements of
the eight characters aabbbbcd. There are
8!/(2!4!1!1!) such unique arrangements.

