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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.
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