Math 375 Final Exam Solutions Spring 2010 Name ... - MavDISK

Math 375 Final Exam Solutions Spring 2010 Name:
Show all work
1. Let an be the number of ternary strings (characters 0, 1, 2) with no
repeated 0s. Find a recurrence relation for the sequence a1, a2, a3, . . . , and
use it to calculate a1 through a8. There is no need to find a formula for an.
Solutions: Strings of length n can begin with 0, 1, or 2. If a string begins
with 0, there are two choices for the next character (1 or 2), and the remaining
n−2 positions must not contain repeated 0s. There are 2an−2 of these strings.
If a string begins with 1 or with 2, the remaining n − 1 positions must not
contain repeated 0s. There are 2an−1 of these strings. So the recurrence
relation is
⎧
⎪⎨ 2an−2 + 2an−1 n ≥ 3
an = 3
⎪⎩
8
n = 1
n = 2.
Therefore the first eight terms of the sequence are
3, 8, 22, 60, 164, 448, 1224, 3344 . . .
1

2. Let an be the number of binary strings with an even number of 0s and no
repeated 1s. Let bn be the number of binary strings with an odd number of 0s
and no repeated 1s. Find a system of recurrence relations for the sequences
a1, a2, a3, . . . and b1, b2, b3, . . . and use the system to calculate a1 through a8
and b1 through b8. There is no need to find a formula for an or bn.
Solution: Strings of length n with an even number of 0s and no repeated
1s must start with 0 or 1. If it starts with 0, it must be followed by an odd
number of 0s and no repeated 1s. Contribution: bn−1. If it starts with 1, it
must be followed by 0 and then an odd number of 0s and no repeated 1s.
Contribution: bn−2. Strings of length n with an odd number of 0s and no
repeated 1s must start with 0 or 1. If it starts with 0, it must be followed
by an even number of 0s and no repeated 1s. Contribution: an−1. If it starts
with 1, it must be followed by 0 and then an even number of 0s and no
repeated 1s. Contribution: an−2. So the system of recurrence relations is
⎧
⎪⎨ bn−1 + bn−2 n ≥ 3
an = 1 n = 1
⎪⎩
1 n = 2
⎧
⎪⎨ an−1 + an−2 n ≥ 3
bn = 1
⎪⎩
2
n = 1 .
n = 2
The first eight terms of the a sequence are
1, 1, 3, 4, 6, 11, 17, 27
and the first eight terms of the b sequence are
1, 2, 2, 4, 7, 10, 17, 28.
2

3. (a) Find a generating function for the recurrence relation an = 2an−1 +5 n ,
a0 = 7.
(b) Extract the coefficient of x 100 in your generating function to compute
a100. Use partial fraction decomposition.
Solution:
g(x) =
g(x) = a0 + a1x + a2x 2 + · · · =
7 + (2a0 + 5)x + (2a1 + 5 2 )x 2 + · · · =
7 + 2xg(x) + 5x
1 − 5x =
2xg(x) +
7 − 30x
(1 − 2x)(1 − 5x)
7 − 30x
1 − 5x ,
= 16
3
1 5 1
+
1 − 2x 3 1 − 5x ,
a100 = [x 100 ] = 16
3 2100 + 5
3 5100 .
3

4. (a) Using the method of roots, find a formula for an given the recurrence
relation an = 2an−1 + n2 n , a0 = 2.
(b) Check your work by computing a1 through a3 in two ways: first using
the recurrence relation, second using your formula for an.
Solution: Rewriting the recurrence relation as
an − 2an−1 = n2 n ,
the homogeneous part contributes a root of 2 and the non-homogeneous part
contributes two roots of 2. So we must have
The sequence begins
an = α · 2 n + β · n · 2 n + γ · n 2 · 2 n .
a0 = 2, a1 = 2a0+1·2 1 = 6, a2 = 2a1+2·2 2 = 20, a3 = 2a2+3·2 3 = 64, · · · .
This yields the system of equations
hence
2 = α, 6 = 2α + 2β + 2γ, 20 = 4α + 8β + 16γ
α = 2, β = 1 1
, γ =
2 2 ,
an = (2 + 1 1
n +
2 2 n2 )2 n .
Check: using n = 3, (2 + 3 9 + 2 2 )23 = 64.
4

5. Using the Principle of Inclusion-Exclusion, find the number of permutations
of the numbers 1 through 9 in which 2 cannot immediately follow
1, 3 cannot immediately follow 2, and 4 cannot immediately follow 3. For
example, 129876543 is bad, 129876534 is bad, and 987123456 is bad, but
987654321 is good.
Solution: There are 3 simultaneous conditions here: no 12 and no 23 and no
34. There are 9! permutations in all. The bad ones are in A1 ∪A2 ∪A3, where
A1 contains permutations with 12, A2 contains permutations with 23, and
A3 contains permutations with 34. The total number of bad permutations is
n1 − n2 + n3. We have
n1 = |A1| + |A2| + |A3| = 8! + 8! + 8!
n2 = |A1 ∩ A2| + |A1 ∩ A3| + |A2 ∩ A3| = 7! + 7! + 7!
n3 = |A1 ∩ A2 ∩ A3| = 6!
Hence the number of good permutations is
9! − 3 · 8! + 3 · 7! − 6! = 256320.
5

Extra Credit Problem 6: Simplify 100
k=0
Solution:
100 k 100−k k3 7 .
k
(x + y) 100 100
100
= x
k
k y 100−k
k=0
(x + 7) 100 100
100
= x
k
k 7 100−k
k=0
100(x + 7) 99 100
100
= kx
k
k−1 7 100−k
k=0
100x(x + 7) 99 100
100
= kx
k
k 7 100−k
k=0
300 · 10 99 100
100
= k3
k
k 7 100−k .
k=0
6

Extra Credit Problem 7: Using a generating function, find the number of
solutions to e1 + e2 + e3 + e4 + e5 = 100, 1 ≤ ei ≤ 25.
Solution: The answer is the coefficient of x 100 in
(x + x 2 + · · · + x 25 ) 5 = x 5 (1 + x + · · · + x 24 ) 5 = x 5 (1 − x 25 ) 5 1
=
(1 − x) 5
x 5 (1 − 5x 25 + 10x 50 − 10x 75 + 5x 100 − x 125 )
∞
n + 4
x
4
n .
The contributions to x100 come from n = 95, 70, 45, 20. Hence the number of
solutions is
99 74 49 24
− 5 + 10 − 10 .
4 4 4 4
7
n=0

Extra Credit Problem 8: Using a generating function, find the number of
ways to place 50 people into 5 distinct buses in such a way that there are an
even number of people in the first, third, and fifth buses.
Solution: The number of ways to place ei people in bus i, 1 ≤ i ≤ 5, is equal
to the number of rearrangements e1 1s, e2 2s, e3 3s, e4 4s, and e5 5s, namely
50! . Therefore the total number of ways to load the bus, summing
e1!e2!e3!e4!e5!
over all the cases, is
50!
e1!e2!e3!e4!e5!
e1, e3, e5 ∈ {0, 2, 4 . . . }
e2, e4 ∈ {0, 1, 2, . . . }
e1 + e2 + e3 + e4 + e5 = 50
.
This equal to the coefficient of x 50 in
x −x e + e
50!
2
which is equal to
3
(e x ) 2 = 50!
8 (e3x − 3e x + 3e −x − e −3x )e 2x =
50!
8 (e5x − 3e 3x + 3e x − e −x ),
550 − 3 · 350 + 2
.
8
8