334
Solutions
to
Exercises
2
1
0
0
1
2
1
0
of
spanning trees is the determinant
of
the
(n

1)
x
(n

1)
matrix
0
1
2
0
0
0
1
2
It
is a straightforward exercise in mathematical induction to prove that this determinant is
n.
(b) There are
n
spanning trees not containing edge
la
(by part (a». Now
suppose
la
is in some spanning tree. We must now delete exactly one
of
the edges 12,23,
...
,
(a

l)a
and exactly one
of
the edges
a( a
+
1),
...
,n1.
There are
aI
choices in the first list and
n

(a

1)
choices in the second. So the number
of
trees containing
la
is
(a

1)(n

(a

1))
and the total number
of
trees is
1
n1J2
...
. 3
.
.
0
••••••
a
n
+
(a

1)(n

(a

1))
=
an

(a

1)2.
(Alternatively, use Kirchhoff's Theorem!)
16. [BB] There are
(~)
possible edges from which we
choose
n

1.
The number
of
graphs is, therefore,
n
No.
of
trees
No.
of
graphs
(~~1).
The number
of
trees on
n
labeled vertices is
nn2.
For
n
::;
6, the table shows the numbers
of
trees
as compared with the numbers
of
graphs.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 any
 Determinant, Graph Theory, Mathematical Induction, Shortest path problem, shortest path

Click to edit the document details