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MATH 239 Assignment 5  Suggested Solutions
1(a)
[1 mark]
The vertices of
A
n
are given by the
n
element subsets of
{
1
,...,
2
n
+ 1
}
, so
the number of vertices in
A
n
is given by
p
=
p
2
n
+ 1
n
P
,
n
≥
1
.
(b)
[1 mark]
For any vertex
α
, the neighbours are given by the
n
element subsets of the set
of
n
+1 elements not in
α
, so
α
has degree
(
n
+1
n
)
=
n
+1,
n
≥
1. Thus
A
n
is an (
n
+1)regular
graph, and from the Handshake Theorem, the number of edges in
A
n
is given by
q
=
1
2
p
(
n
+ 1) =
n
+ 1
2
p
2
n
+ 1
n
P
,
n
≥
1
.
(c)
[2 marks]
Suppose that a triangle exists in
A
n
, with vertices
α,β,γ
. Then
α,β,γ
are
all disjoint from each other (i.e.,
α
∩
β
=
α
∩
γ
=
β
∩
γ
=
∅
). Then
S
=
α
∪
β
∪
γ
is a subset
of
{
1
,...,
2
n
+ 1
}
with

S

= 3
n
, which implies that 3
n
≤
2
n
+ 1. Subtracting 2
n
from each
side of this inequality, we get
n
≤
1. We conclude that
A
n
cannot contain a triangle if
n >
1.
When
n
= 1,
A
n
does indeed contain the triangle
{
1
}
,
{
2
}
,
{
3
}
.
2(a)
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 Spring '08
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