MATH 460 2002/2003
Assignment 4
Due January 27, 2003
1. In how many ways can three 0s, three 1s, and three 2s be arranged so that no three
adjacent digits are the same in an arrangements?
Solution: Let ai be the property that three is are together (i = 0,
Ch
R
3. 1
U
s i n
c
Th
p
e
b
x
o
l e m
a
f
p
e
t o
i s
= 8
3
,
n 1
t h
h
a v
e
a
e
s i n
l s
d
a
o
t
e n
t h
o v a
l s
a
S
p
2
r r e n
c e
e
o
n
l e t
c e
d
i
t h
w
F
e
)
r e c u
c e
=
s t i c k
e r e n
e
( x
w
f o
n=
t h
r r e n
r
c e
i t h
e
a
a
0
g
e
MA 460 2000/2001 Final Exam
April 16, 2001
Dr. B. Zhou
Name
Instructions. All answers should be clear and complete. Show all your works to justify your
answers. Partial credit will be given for the part of your work that leads to a correct answer. You
may
MATH 460 2001/2002 Midterm Test
December 12, 2001
Dr. Bing Zhou
Name
Instructions. All answers should be clear and complete. Show all your works. Partial
credit will be given for the part of your work that leads to a correct answer. You may leave
notati
MATH 460 2002/2003
Assignment 5
Due March 3, 2003
1. (a) Show that if T is a tree with n vertices, then the independence number of T , (T ), satises the
inequality
(T ) n/2 .
Is there a tree T such that (T ) = n/2 ?
(b) Extend the result in (a) as much a
E
x
1
p
o n
e n
. 2
T
E
h
o
t i a l
e
g
x
e n
g e n
p
o
e r a t i n
n
e r a
e
n
t i n
g
f u
n
t i a
f u
g
l
c t i o
n
g
c t i o n
e
n
e
n
s
1
r a
o
t i n
f
t h
g
e
f u
p
n
c
e r m
u
t i o
n
s
t a t i o
n
s
f o
r
a
s e t
o f
t w
o
e l e m
e n
t s
a
, b
i n
C
G
1. 1
L
O
e t
0
r d
A
,
i n
=
r y
( a
, . . .
1
a
0,
e
f u
b
g
a
n
e
r a
, . . . , a
1
n
e
c t i o
n
E
t i n
r,
s
c a l l e d
a
g
e n
e r a
t i n
( t h
g
f u
E
e
n
b
d
c t i o
a
i c a
= a
0
n
t h
E
x
e
a
o
m
r d
p
i n
a
r y
g
e n
e r a t i n
1. 1. 1 W
MATH 460 2002/2003
Assignment 2
Due Nov. 6, 2002
1. Let S (n, r) be the Stirling number of the second kind.
(a) Evaluate S (1, r) and S (2, r) .
Solution:
S (1, r) =
S (2, r) =
=
=
S (n, r) =
1
n!
i n
i=0 (1) i (n
Pn
i)r
1
1
1 X
r
(1)i
(1 i) = 1 0 = 1
evah eW . a
01
.x
99 = 1
1
1
11
001 = 11
=
+
2
1
x x
+
x
001 = 11
+
2 1 =
9
01
3
+1
si ix hcae rof noitcnuf gnitareneg ehT
x
x
+ 1
dna
0
= 01
1
001
2
k
+ 2
,. , , i , ix
x , a a x ,. , a a x , a x teL : noituloS
. ,. , , , cfw_ tes eht morf sregetni fo
Ch
I N
4. 1
B
a
De
n
i t i o
t h
f o
s i c
r m
o
e
f
t h
s e t
j o
e
o
i n
r a
j a
h
n
x
r a
W
e
a
m
p
m
h
a
n
O
x
s e
t i o n
i n
s
V
y
o f
cfw_
e
n
N
g
a
w
t e r
T
O
4
G
t i o
n
R
p
, y
A
g e ,
P
H
b
e
s a y
j a
c i d
c
a
x
x
, a
y
A
c o
m
e d
p
p
l a