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200
Solutions to Exercises
(c) Let
Al
be the set of solutions to (*) with
Xl
~
8. Define
A2, A3
and
A4
similarly. The question
asks for the number of elements in
Al C
n
A2
c
n
A3
c
n
A4
c
=
(AI
U
A2
U
A3
U
A4)C
(by one of
the laws of De Morgan). By the Principle of InclusionExclusion,
IA
l
UA
2
UA
3
UA
4
1
=
L
IAil
L
IAi
n
Ajl
+
L
IAi
n
Aj
n
AklIAl
n
A2
n
A3
n
A41
i<j
i<j<k
By part (a),
JAIl =
560. Similarly, each of the four terms in
Ei IAil
is 560. By part (b),
IA
l
nA
2
1
=
56 and similarly, each of the
@
=
6 terms in
Ei<j IAinAjl
is 56. Now
AlnA2
nA3
is the set of solutions to (*) with Xl,
X2, X3
~
8 and
z
~
o. There are no such solutions! Thus,
IAI U
A2
U
A3
U
A41
=
4(560) 
6(56)
=
1904. Since the number of solutions to (*) in
nonnegative integers is
(~f)
=
2024, the number of solutions with each variable at most seven is
2024 
1904
=
120.
(d) We proceed as in part (c). Let
Ai
be the set of all solutions to the given equation in which variable
Xi
~
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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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