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170
Solutions to Exercises
(a) This part asks us for
IA U
B
U C U DI
and we have
IA U
B
U C U DI
=
IAI
+
IBI
+
ICI
+
IDI
 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
=
3333
+
2000
+
1428
+
909 
666  476 
303 
285 
181 
129
+
95
+
60
+
43
+
25 
8
=
5845.
(b) This part asks for
I(A
n
B)
'. (C U
D)I
=
IA n BI
I(A
n
B)
n (C U
D)I·
Now
(A
n
B)
n (C
U
D)
=
(A
n
B
n C)
U
(A
n
B
n
D),
so
I(A
n
B)
n (C
U
D)I
=
IA n
B
n
CI
+
I(A
n
B
n
D)II(A
n
B
n C) n
(A
n
B
n
D)I
=
IAnBnCl
+
I(AnBnD)II(AnBnCnD)1
= 95
+
60  8 = 147.
The answer is 666  147 = 519.
(c) This question asks for the number of elements in
((AnBnC) '.D)
U
((AnBnD) '.C)
U
((AnCnD) '.B)
U
((BnCnD) '.A).
Since these sets are
pairwise disjoint,
that is, the intersection of two of the four is empty, the
number of elements in the union is the sum of the elements in each set. Since
I(A
n
B
n CI
'. DI = IA
n
B
n
CIIA
n
B
n C n DI
=
95  8
=
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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