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Math 55: Discrete Mathematics
UC Berkeley, Spring 2009
Solutions to Homework # 6 (due March 9)
§
5.1, #8: There are 26 choices for the ﬁrst letter, then 25 for the second and 24
for the third to avoid repetition. Thus there are 26
·
25
·
24 = 15600
possibilities.
§
5.1, #16: We separate into disjoint cases and apply the Sum Rule. Either the
ﬁrst letter is
x
(leaving the others unconstrained) or the ﬁrst is not
x
but the second is
x
, or the ﬁrst two are not but the third is, or the ﬁrst
three are not but the fourth is.
Thus there are 26
3
+25
·
26
2
+25
2
·
26+25
3
= 66351 diﬀerent such strings.
(Note: “
x
is the ﬁrst letter” and “
x
is the second letter” are
not
disjoint
cases!)
§
5.1, #19: a) Should be
b
100
7
c  b
49
7
c
= 7 multiples of 7 between 50 and 100; these
are
{
56
,
63
,
70
,
77
,
84
,
91
,
98
}
.
b)
b
100
11
c  b
49
11
c
= 5;
{
55
,
66
,
77
,
88
,
99
}
.
c) The numbers that are both multiples of 7 and 11 are precisely the
multiples of 77; clearly 77 is the only one of these between 50 and 100.
§
5.1, #24: a) 10
·
9
·
8
·
7 = 5040
b) 10
3
·
5 = 5000
c) Since the non9 digit can be in one of four places (and these cases
are disjoint) and can be any of the other 9 digits, there are 4
·
9 = 36
such strings.
§
5.1, #30: a) 26
8
b) 26
·
25
·
24
·
23
·
22
·
21
·
20
·
19
c) 26
7
d) 25
·
24
·
23
·
22
·
21
·
20
·
19
e) 26
6
f)26
6
g) 26
4
h) 26
6
+ 26
6

26
4
, by the InclusionExclusion Principle.
1
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5.1, #55: You can draw a tree graph in which each branching asks whether a given
element belongs to the subset, each node at a given level is labeled with
the sum of its elements thus far, and only nodes whose sum of elements
is
≤
28 are included. The graph (too unwieldy to picture here) indeed
has 17 nodes at the bottom, corresponding to the 17 such subsets.
§
5.2, #9: There must be 99
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