332
Solutions to Exercises
3
1
0
0
0
1
1
2
1
0
0
0
The
value
of
the
(1,1)
cofactor is
0
1
3
1
0
1
=
65.
0
0
1
2
1
0
0
0
0
1
3
1
1
0
1
1
1
4
We conclude that there are
65
spanning trees.
3.
lC
n
has
nn2
spanning trees each with
n

1 edges. Hence, the total number
of
all edges used in all
spanning trees is
(n

l)nn2.
Now each
of
the
(~)
edges in
lC
n
is equally likely to
be
included in a
spanning tree. Hence, the number
of
spanning trees containing
e
is
(n

(i)nn2
=
2nn3.
4.
[BB] By Theorem
12.2.3,
the numbers are
1
1
=
1,
2°
=
1,
3
1
=
3,
4
2
=
16,
53
=
125,
and
6
4
=
1296.
5. [BB]
A
BA BA
BA
BA
BA BA BA
B
:J~nXClXUX
C
DC DC DC DC DC DC DC
D
A
BA BA BA BA
BA BA BA
B
NSZV1~/1k:~
C
DC
DC DC DC DC DC DC
D
6.
lC
7
(labeled) has 7
5
spanning trees. (See Theorem
12.2.3.)
7. (a) [BB]
lC
2,2
has four spanning trees (obtained by deleting each edge in succession). They are all
isomorphic to
0
0
0
0 .
(b)
lC
2,3
has
12
spanning trees as shown.
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 Summer '10
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 Graph Theory, Isomorphism, Spanning tree, dc dc dc, BA BA BA B A BA BA B

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