CSCE 896 Graph Theory Homework 3
Fall 2010
1. Let
e
be an edge of
K
n
. Use Cayley’s Theorem to prove that
K
n

e
has (
n

2)
n
n

3
spanning trees.
Proof:
Let
V
(
K
n
) =
v
1
,v
2
,......
,v
n
. We may assume that
e
=
v
1
v
2
, and we count the
number
N
of spanning trees which contain the edge
e
. Let
T
be such a tree. Then
deg
T
(
v
1
) +
deg
T
(
v
2
) =
k
+ 2 for some integer
k
≥
0. If
k
= 0, then
n
= 2, and the
statements proven. Let
k >
1, given such a
k
, let’s assume there are
a
k
number of trees
T
that have
deg
T
(
v
1
) +
deg
T
(
v
2
) =
k
+ 2.
Let
T
0
be obtained from
T
by replacing the pair of vertices
{
v
1
,v
2
}
by a new vertex
u
and joining
u
to all neighbors of
v
1
and
v
2
in
T
. Then
T
0
is a tree of order
n

1
≥
2 and
deg
0
T
(
u
) =
k
≥
1. It is clear that
T
7→
T
0
deﬁnes a 2
k

to

1 map from the set of trees
on
{
v
1
,...,v
n
}
with
deg
T
(
v
1
) +
degT
(
v
2
) =
k
+ 2, to the set of trees on
{
u,v
3
,......
,v
n
}
with
degT
0
(
u
) =
k
, since every vertex
v
i
,
3
≤
i
≤
n
, of
T
can be connected to either
v
1
or
v
2
, but not both.
We know that there are
a
k
= 2
k
(
n

3
k

1
)(
n

2)
n

k

2
, and
N
=
n

2
X
k
=1
a
k
=
n

2
X
k
=1
2
k
(
n

3
k

1
)(
n

2)
n

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 Fall '08
 STAFF
 Graph Theory, Planar graph, Glossary of graph theory, Complete bipartite graph

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