Math 55: Discrete Mathematics, Fall 2008
Homework 12 Solutions
7.5: 8. 270

64

94

58 + 26 + 28 + 22

14 = 116
* 14.
26!

23!

23!

24! + 21!.
In this problem our three sets are
A
ﬁsh
=
{
strings containing
ﬁsh
}
, and
A
rat
,
A
bird
deﬁned similarly. We get

A
ﬁsh

= 23! by treat
ing
ﬁsh
as a single letter, and the others likewise. We get

A
ﬁsh
∩
A
rat

= 21! by treating both
ﬁsh
and
rat
as single letters. The other intersections are empty, because
bird
has letters in
common with both
ﬁsh
and
rat
.
20. 5
·
10
4

(
5
2
)
10
3
+
(
5
3
)
10
2

(
5
4
)
10 +
(
5
5
)
1. By the binomial theorem, this is equal to
(1

10)
5
+ 10
5
= 10
5

9
5
= 40951.
7.6: 6. The nonsquarefree positive integers
<
100 are divisible by 2
2
, 3
2
, 5
2
or 7
2
. Let
A
p
be the set of those divisible by
p
2
. Note that
A
p
∩
A
q
=
A
pq
since
p
and
q
are relatively
prime, and likewise for products of more than two factors. Moreover, if the product is
≥
10,
then the intersection is empty, so the only nonempty intersection that contributes to the
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This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at Berkeley.
 Fall '08
 STRAIN
 Math

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