This preview shows page 1. Sign up to view the full content.
332
Solutions to Exercises
3
1
0
0
0
1 2 1
0
0
0
The value of the (1,1) cofactor is
0
1 3
0
=
65.
0
0
0
0
0
0
0
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
NSZV1~/1k:~
6. lC
7
(labeled) has 7
5
spanning trees. (See Theorem 12.2.3.)
7. (a) [BB]
2,2
has four spanning trees (obtained by deleting each edge in succession). They are all
isomorphic to
0
0
0
0 .
(b)
2,3
has 12 spanning trees as shown.
The graphs in the top row form one isomorphism class and those in the bottom row a second.
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details