Chapter 6
181
Chapter 6 Review
1. By the Principle ofInclusionExclusion,
IA n BI
=
IAI
+
IBI
IA UBI,
so
IA n BI > O.
2. Let
A, B,
C,
D
be the sets of integers between 1 and 2000 (inclusive) which are divisible by 2, by 3,
by 5, and by 7, respectively. We want
I(A
U
B
U C U
D)I.
Now
IA U
B
U C U DI
=
IAI
+
IBI
+ ICI +
IDI
We have
 IA n BI  IA n CI  IA n DI  IB n CI  IB n DI  IC n DI
+ IA n
B
n CI + IA n
B
n DI + IA n C n DI + IB n C n DI
IAnBnCnDI·
IAI
=
l
20
2
00
J
=
1000,
IBI
=
l20
3
00
J
=
666,
ICI
=
l
20
2°
J
=
400,
IDI
=
l20~O
J
=
285,
IA n BI
=
l20
6
00
J
=
333,
IA n CI
=
l2~gO
J
=
200,
IB n CI
=
l2~~O
J
=
133,
IB n DI
=
l2~~O
J
=
95,
IA n DI
=
l2~~O
J
= 142,
IC n DI
=
l2~~O
J
=
57,
IA n
B
n CI
=
l2~gO
J
=
66,
IA n
B
n DI
=
l2~~O
J
=
47,
IA n C n DI
=
l2~gO
J
=
28,
IB n C n DI
= l2
I
0
0
0
5
0
J
=
19,
IA n
B
n C n DI
= l2
2
0
I
O
O
O
J = 9.
SO
IA U
B
U C U DI
= 1542.
3. (a) A palindrome is a word which is spelt the same way forwards as backwards, such as
SUNUNUS.
(b) We subtract from the total number of sevenletter palindromes which begin with S the number
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 Summer '10
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 Graph Theory, Integers, Sets, ABBA, IAI

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