MATH 681
Final Exam
Answer exactly four of the following six questions.
Indicate which four you would like
graded!
Binomial coefficients, Stirling numbers, and arithmetic expressions need not be simplified
in your answers.
1.
(12 students attempted this problem)
Answer the following questions about find
ing the number of words
a
n
of length
n
with letters A, B, and C using the letter “A”
at least once.
(a)
(5 points)
Find an exponential generating function for
a
n
.
The exponential generating function associated with a string of one or more A’s
is:
x
+
x
2
2
+
x
3
6
+
x
4
24
+
· · ·
=
e
x

1
The exponential generating function associated with a string of zero or more B’s
(or zero or more C’s) is:
1 +
x
+
x
2
2
+
x
3
6
+
x
4
24
+
· · ·
=
e
x
Multiplying these three together, a mixed string of one or more A’s, zero or more
B’s, and zero or more C’s is associated with the exponential generating function:
(
e
x

1)(
e
x
)(
e
x
) =
e
3
x

e
2
x
(b)
(5 points)
Develop an argument to show that
a
n
satisfies the recurrence
a
n
=
2
a
n

1
+ 3
n

1
with
a
0
= 0
.
Clearly
a
0
= 0, since the unique string of length zero has no “A” in it. We may
use a casewise argument on the first character (o last character, if you prefer) in
a string of length
n
.
If the first character is “B” or “C”, than the substring consisting of the
n

1
subsequent characters must be a string of A’s, B’s, and C’s which contains at least
one A: in other words, a string of length
n

1 satisfying our
original
criterion.
We know there are
a
n

1
such strings by definition, so there are 2
a
n

1
strings of
length
n
beginning with “B” or “C”.
If, on the other hand, the first character is an “A”, then the
n

1 subsequent
characters are no longer restrained — the string as a whole must contain an A,
but it is guaranteed to do so! Thus, the following
n

1 characters can be any
string of “A”, “B”, and “C”, which we know can be produced in 3
n

1
ways.
Putting these two cases together, we see that
a
n
= 2
a
n

1
+ 3
n

1
.
(c)
(5 points)
Using a method of your choice, find a closedform expression for
a
n
.
It is easy to derive this from the exponential generating function found in part
(a), and only slightly harder to derive it from the recurrence relation found in
part (b). There is also a direct enumeration using exclusion methods. Here the
EGF approach is demonstrated.
Page 1 of 6
December 12, 2009
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MATH 681
Final Exam
Expanding the exponential generating function we already discovered:
∞
X
n
=0
a
n
x
n
n
!
=
e
3
x

e
2
x
=
∞
X
n
=0
(3
x
)
n
n
!

∞
X
n
=0
(2
x
)
n
n
!
=
∞
X
n
=0
(3
n

2
n
)
x
n
n
!
so
a
n
= 3
n

2
n
.
2.
(6 students attempted this problem)
Prove the following identities using combi
natorial arguments, where
F
n
represents the Fibonacci sequence indexed with
F
0
= 1
,
F
1
= 1
,
F
2
= 2
.
(a)
(5 points)
k
(
n
k
)
=
n
(
n

1
k

1
)
.
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 Fall '09
 WILDSTROM
 Binomial, Order theory, Recurrence relation, Generating function, exponential generating function, Dilworth

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