Math 5366
Fall 2015
Exam 2
Solutions
1. (20 points) Find the appropriate generating function (in closed form) for each of the
following problems. Do not find the coefficient of xn .
(a) In how many ways can n balls be put in 4 boxes if each box is nonempty?
x
.
1−x
x4
The generating function for the problem is the fourth power of this,
.
(1 − x)4
Solution: The generating function for each box is x + x2 + x3 + · · · =
(b) How many quaternary sequences (0’s, 1’s, 2’s, 3’s) of length n are there having at
least 1 of each of the four digits?
Solution: Here, the generating function is (x +
x2
2!
+
x3
3!
+ · · · )4 = (ex − 1)4 .
(c) In how many ways can n balls be put in 4 boxes if the first box has an even number
of balls and the last box has an odd number of balls?
1
Solution: The generating function for the first box is 1 + x2 + x4 + · · · = 1−x
2.
x
3
5
For the last box, it is x + x + x + · · · = 1−x2 . For the problem we have
1
1
x
x
1
=
.
1 − x 2 1 − x 1 − x 1 − x2
(1 − x2 )2 (1 − x)2
(d) How many quaternary sequences of length n have an even number of 0’s and an
odd number of 1’s?
ex + e−x
ex − e−x
, odd number of 1’s:
, our
2
2
ex + e−x ex − e−x 2x e4x − 1
generating function is
e =
.
2
2
4
Solution: Even number of 0’s:
2. (20 points) Find the generating function for each of the following problems. Then
SOLVE for the number asked.
(a) In how many ways can n balls be put into 4 boxes if the first box has at least 2
balls?
Solution: The generating function for the first box is x2 +x3 +x4 +· · · =
x2
.
1−x
x2
for the problem, the generating function is
, and the coefficient of xn in
4
(1
−
x)
n−2+4−1
n+1
this is
=
.
4−1
3
(b) How many quaternary sequences of length n have at least two 0’s?
2
3
Solution: For 0 we need x2! + x3! + · · · = ex − 1 − x. The rest just contribute ex
each. The generating function for the problem is e4x − e3x − xe3x . The answer
to the question is n! times the coefficient of xn in this expression or
n
3n
3n−1
4
−
−
n!
= 4n − 3n − n3n−1 .
n!
n! (n − 1)!
3. (10 points) Find a recurrence, with initial conditions, for an , the number of ternary
sequences of length n in which all the 0’s occur before any 2. Hint: After the first 2
appears, the remaining digits can only be 1’s and 2’s.
Solution: If the sequence starts with either 0 or 1, then filling in the remaining
n − 1 positions is the same kind of problem: they must have all 0’s occurring before
any 2. The count for this is an−1 , so there are 2an−1 strings of this type. If the
string starts with a 2, then the remaining n − 1 positions consist of only 1’s and 2’s,
and there are a total of 2n−1 strings of this type. The total count is an = 2an−1 +2n−1 .
For the initial conditions, the empty string fulfills the requirements so a0 = 1. Alternatively, all strings of length 1 are ok so a1 = 3.
As a check, the recurrence continues 1, 3, 8, 20, 28, . . . . How many bad strings
of length 3 are there? I count 7: 200, 201, 202, 220, 210, 020, 120. This leaves
33 − 7 = 20 good strings, as it should.
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4. (20 points) Solve each of the following recurrence relations
(a) an = 7an−1 − 10an−2 , a0 = 4, a1 = 11.
Solution: The characteristic polynomial is x2 − 7x + 10 = (x − 2)(x − 5) so
the general solution is an = A2n + B5n . The initial conditions give us A + B =
4, 2A + 5B = 11. Multiplying the first by 2 and subtracting from the second,
3B = 3. Thus, B = 1, A = 3, and the solution is an = 3 · 2n + 5n .
(b) an = 6an−1 − 9an−2 , a0 = 1, a1 = 9.
Solution: Here, the characteristic polynomial is x2 − 6x + 9 = (x − 3)2 . Since
there is a multiple root the general solution is an = (An + B)3n . The initial
conditions now say B = 1, 3A + 3B = 9, or A = 2. The solution is an =
(2n + 1)3n .
5. (20 points) solve each of the following inhomogeneous recurrence relations. Hint: You
solved the homogeneous problem in part 3a. Sorry about that–it should have been 4(a).
(a) an = 7an−1 − 10an−2 − 2 · 3n , a0 = 12, a1 = 39.
Solution: The general solution to an inhomogeneous recurrence has the form
(Particular solution to inhomogeneous recurrence) + (general homogeneous solution). We know the general homogeneous solution is A2n + B5n from question 4. For a particular solution, we should guess an = P 3n . Trying this,
P 3n = 7P 3n−1 − 10P 3n−2 − 2 · 3n . Dividing by 3n−2 , 9P = 21P − 10P − 18, or
2P = 18. So the general solution to the recurrence is A2n + B5n + 9 · 3n . The
initial conditions yield A + B + 9 = 12, 2A + 5B + 27 = 39. From these, we get
B = 2, A = 1 so the solution we seek is an = 2n + 2 · 5n + 3n+2 .
(b) an = 7an−1 − 10an−2 + 3 · 5n−1 , a0 = 1, a1 = 7.
Solution: Since 5n is already part of the homogeneous solution, for a particular
solution we should guess an = P n5n . Trying this, P n5n = 7P (n − 1)5n−1 −
10P (n − 2)5n−2 + 3 · 5n−1 . Dividing by 5n−1 , 5nP = 7(n − 1)P − 2P (n − 2) + 3.
If you really trust yourself, you can plug in a value for n to get P. One student
use n = 2 to get 10P = 7P + 3. I just worked with the original expression:
5nP = 7nP − 7P − 2nP + 4P + 3 → nP (5 − 7 + 2) + P (7 − 4) = 3.
Either way, P = 1, and the general solution is A2n + B5n + n · 5n . The initial
conditions give A + B = 1, 2A + 5B + 5 = 7. From these, we get A = 1, B = 0.
The solution we seek is an = 2n + n5n .
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Here is another approach to problems like number 5. Take part (a). If we know that
the particular solution has to have the form P 3n and the general solution will be an =
A2n +B5n +(particular) = A2n +B5n +P 3n , then we can create another initial condition
and solve for A, B, P using those. In this case, a2 = 135. We have
A + B + P = 12,
2A + 5B + 3P = 39,
4A + 25B + 9P = 135.
Subtracting twice the second from the third and then twice the first from the second,
15B + 3P = 57, 3B + P = 15, and we get 6B = 12 → B = 2, P = 9, A = 1.
On 5(b), we could not use an = A2n + B5n + P 5n because 5n would be used twice so we
need an = A2n + B5n + P n5n . With a2 = 54,
A + B = 1,
2A + 5B + 5P = 7,
4A + 25B + 50P = 54.
We quickly find A = 1, B = 0, P = 1. This method requires that we know the form of
the particular solution. It usually requires more effort than the method outlined in the
book and in class (more variables and more equations to solve) but in small cases it can
be pretty fast.
6. (10 points) Solve each of the following recurrence relations. You may use any approach
that can be sufficiently justified. I will give extra credit if you use generating functions.
(a) an = an−1 + 3an−2 + 9an−3 + · · · + 3n−1 a0 , a0 = 3.
Solution: Here are two approaches: First, if we just plug in values, a1 = a0 =
3, a2 = a1 + 3a0 = 12, a3 = 12 + 3 · 3 + 9 · 3 = 48, we might be able to guess
that an = 3 · 4n−1 , when n ≥ 1, with a0 = 3, which technically does not fit the
pattern. With this guess, we could verify by induction that the guess is right:
if true through n − 1 then
an = 3 · 4n−2 + 32 · 4n−3 + · · · + 3n−1 40 + 3n .
All but the last term form a geometric progression (ratio 43 ) so
3 · 4n−2 − 43 3n−1 40
an = 3 +
= 3n + 3 · 4n−1 − 3n = 3 · 4n−1 ,
1 − 43
n
as desired.
More thematic is the following: If n ≥ 2 then we can write an−1 = an−2 +3an−3 +
· · · + 3n−2 a0 . We multiply this by 3 and subtract from the original recurrence:
an = an−1 + 3an−2 + 9an−3 + · · · + 3n−1 a0
−(3an−1 =
3an−2 + 32 an−3 + · · · + 3n−1 a0 )
to get an − 3an−1 = an−1 , or an = 4an−1 , but not starting at n = 1 but at n = 2.
Since a1 = 3, we have, for n ≥ 1, an = 3 · 4n−1 , as above.
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(b) an = nan−1 + 3n − 3, a0 = 1. Strong hint: One approach is to guess a particular
solution of the form an = P, and then solve the homogeneous problem.
Solution: The point of this problem is to stress this idea that for an inhomogeneous system,
General solution = General homogeneous solution + Particular solution.
The homogeneous problem, an = nan−1 can easily be solved by iteration: a1 =
1 · a0 , a2 = 2a1 = 2 · 1 · a0 , a3 = 3 · 2 · 1a0 , etc. so an = An!. But what
about the particular solution? The hint says to try an = P, a constant. We get
P = nP + 3n − 3, or n(P + 3) − (P + 3) = 0, and this should be 0 for all n. This
happens when P = −3 so our particular solution worked. The general solution
is an = An! − 3, and a0 = 1 requires A = 4. The solution is an = 4n! − 3.
7. Some extra credit questions, if you are interested. Generating functions help on (d).
(a) (2 points) Solve the recurrence in question 3.
(b) (8 points) Find, with initial conditions, a recurrence for an , the number of sequences
of length n made of 0’s, 1’s, 2’s, 3’s, 4’s, if all 0’s appear before any 2 and all 2’s
appear before any 4. Solve the recurrence. I will give credit for sufficient progress.
(c) (4 points) Solve the recurrence an = an−1 + 2an−2 + 3an−3 + · · · + na0 , a0 = 1.
(d) (4 points) Solve the recurrence an = nan−1 + (n − 1)an−2 + · · · + a0 , a0 = 1.
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