
Chapter 2 Section 2
(pp 2829)
 #2: disjoint sets: I and II; I and III; I and
IV; II and III; II and IV; III and IV; IV and VI
 #3: I, II, III; I, II, IV; I, III, IV; II, III,
IV; I, II, III, IV
 #4: {f,j,k,l,q,v,w,z}
 #5: 10 choices
 #6: 8 choices
 #11: 17 (5+6+2+4)
Chapter 2 Section 3 (pp 3435)
 #2: 5 * 6 * 2 * 4 = 240 ways
 #3: Look at the most restrictive portion first:
the units digit of the number. It must be odd, so
there are 5 choices for that position. Because the
digits in the number must be unique, there remain
9 digits to choose from. We fill the remaining
three spots (tens, hundreds, thousands positions)
in 9 * 8 * 7 = 504 ways. So, in all there are 5 *
(9 * 8 * 7) = 2520 integers.
 #5: Begin by determining how many 7digit
numbers there are that have unique digits and no
0, then remove from this total those that contain
5 and 6 adjacent. There are 9*8*7*6*5*4*3 7digit
numbers that have unique digits and no 0. Of
these, we cannot have 5 and 6 adjacent. The digits
5 and 6 can be adjacent as ...56... and as
...65.... Also, within any 7digit number, a 56
package can appear in any one of six positions in
the number (first/second, second/third, and so
on). So we have 2*6 = 12 ways/places 5 and 6 may
appear adjacent. With each of these 12 ways,
however, we must consider options for all the
other digits in the 7digit number. We have
already used 5 and 6, so we have 7 available
digits with which to build the rest of the number,
containing 5 open spots for digits. We can fill
those spots in 7*6*5*4*3 ways/ Therefore, the
total restricted 7digit numbers is (12) *
(7*6*5*4*3). Therefore, the number of desired
7digit numbers is (9*8*7*6*5*4*3)  [(12) *
(7*6*5*4*3] = 151,200.
 #7: Each person can choose from 5 shirts. The
desired number of 3shirt sets is 5*5*5 = 125.
 #8: Consider the 5 consonants. There are 5! =
5*4*3*2*1 ways to arrange these. With each
arrangement of the consonants, we have six places
into which we can drop vowels and meet the
requirement of vowel nonadjacency:
___ C___M___P___T___R___
We choose three of these six available
spots and permute the vowels there. There are C(6,3)
* 3! = P(6,3) such ways. In all, then, there are
(5!) * (P(6,3)) = 14,400 ways.
Chapter 2 Review (pp 3538)
 #2: 96 responses
 Because there are five categories of STEM
professionals, The surveyor could get 19
responses from each of the five groups and still
not have 20 from a group. When the surveyor gets
that 96th response, however, it will assure
there are at least 20 responses from the same
group.
 #4:
 (a) 20 choices, by the Addition Principle.
 (b) 96 pairs: using the multiplication
principle, the pair each of the 8 coffee choices
with each of the 12 tea choices. We disregard
order, as stated in the problem
 #7:
 (a) There are two choices for each of the six
dots, either raised or not raised, so there are
2^{6} = 64 different raiseddots
options. One of those, however, is no raised
dots, which is not an option. Therefore, there
are 63 symbols that can be represented.
 (b) Even numbers are 0, 2, 4, 6, and so on.
Here, 0 raised dots is not an option. So for 2,
4, and 6 raised dots, we use combinations to
determine the number of ways to select dots for
raising. We have C(6,2), C(6,4), and C(6,6).
Because these are disjoint (we cannot have
exactly two raised dots and at the same time
have exactly four raised dots), we add these
results: C(6,2) + C(6,4) + C(6,6) = 31 symbols
with an even number of raised dots.
 (c) There are at least a couple ways to do
this.
 The first method is to look at options that
meet the restrictions. In each row, there must
be one or no raised symbols. This can be done
in 3 ways for each row: only left dot raised,
only right dot raised, or neither dot raised.
Because there are three rows with this
required characteristic, there are 3*3*3 = 27
different options. But one of these
optionsno dot raised in any rowisn't
allowed, so there are 26 symbols, in all that
are possible under this restriction.
 A second method is to start with all
possible symbols (63, from part (a) here) are
remove those that cannot be used. Here, those
that cannot be used include any circumstances
with both dots raised in one or more rows. We
need to look at cases here.
 We could have 2 dots raised in every row.
There is only one way to do that.
 We could have two dots raised in two of
the three rows. There are 3 ways [C(3,2)] to
choose exactly two rows at a time, and for
each of these options, there is one way to
raise two dots in each of the two rows and
three ways to raised less than two dots in
the remaining row. Thus, for each tworow
pair here, there are 1*1*3 = 3 symbols, and,
as stated, there are three 2row options for
this. Thus, There are 9 symbols for this
case.
 The last case is two raised dots in
exactly one row. Using similar reasoning as
the previous case, there is one way to raise
two dots in that required row and three ways
to choose less than two dots in the other
two rows. Each of these is possible for each
of the three ways we can select one row for
the two dots. So, for this last case, there
are (1*3*3)*C(3,1) = 9*3 = 27 symbols.
 Add the three cases (disjoint outcomes),
there are 1 + 9 + 27 = 37 ways to have one
or more rows with two raised dots. We
subtract 37 from 63 to show, once again,
that there are 26 symbols with no more than
one raised dot in any row.
Chapter 3 Section 1 (pp 4550)
 #1: 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
 #2: n! = n*(n1)*(n2)*...*2*1
 #6: 10!/(8!2!) = (10*9)/2 = 45
 #8: n!/4! = n*(n1)*(n2)*...*6*5
 #13: (a) P(8,5) = 6720 (b) 8^{5} =
32,768
 #15: (a) 6! * 8! = 29,030,400 (b) (7*6) * (6*5)
* 9! = 457,228,800 (c) 7! * 6! = 3,628,800
 #18: (a) 8! = 40,320 (b) 6! = 720 (c) (4*3) * 6!
= 8640
 #23: (5*6*2*4) * 4! = 5760

