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MATH 239/249 Introduction to Combinatorics
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所属分类：数学mathematics
标签： MATH 239
Introduction to Combinatorics

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MATH 239/249

Introduction to Combinatorics
Hello World, and Thanks!
The following parts of these notes are still UNDER CONSTRUCTION.
(Estimated completion date: Fall 2021.) Most of these are additional topics
not covered during the semester. The old notes are also available on the
course website.
• Proof of Theorem 4.18 (inhomogeneous linear recurrence relations)
• Section 8.4 (planar duality)
• Part III: Chapters 12, 13, and 14 will appear in three installments over
the course of the next year.
Department of Combinatorics and Optimization

Preliminaries.
MATH 239 is an introduction to two of the main areas in combinatorics –
enumeration and graph theory. MATH 249 is an advanced version of MATH
239 intended for very strong students. These courses are designed for stu-
dents in the second year of an undergraduate program in mathematics or
computer science.
The prerequisites required from first-year mathematics are as follows.
• From MATH 135. Abstract algebra I: sets and propositional logic, proofs,
mathematical induction, modular arithmetic, complex numbers, the
Fundamental Theorem of Algebra.
• From MATH 136. Linear algebra I: systems of linear equations, Gaus-
sian elimination, matrix algebra, vector spaces.
• From MATH 137. Calculus I: algebra with power series, open/closed
sets, continuous functions, differentiation (but not integration).
We use the following standard notation for various number systems.
natural numbers N = {0, 1, 2, 3, ...} including zero 0
integers Z = {...,−2,−1, 0, 1, 2, ...}
rational numbers Q
real numbers R
complex numbers C
integers (modulo n) Zn = {[0], [1], [2], ..., [n− 1]}
finite field of prime size Fp = Zp
The cardinality (size) of a set S is denoted by |S|.
For convenience, we use LHS and RHS as shorthand for “left-hand side”
and “right-hand side”, respectively.
9

Part I
Introduction to Enumeration
11

Overview.
Suppose I pay $5 for a lottery ticket – what is the chance that I win a share
of the top prize? It depends on the details, of course. There are a certain
number of ways to win, and a certain number of ways to lose. Enumeration
is the art and science of figuring out this kind of thing. This is the subject of
the first part of these course notes.
There are two broad principles of the subject. The combinatorial ap-
proach is to construct explicit one-to-one correspondences between sets to
show that they have the same size. The algebraic approach is to translate the
information about the problem from combinatorics into algebra, and then to
use algebraic techniques to determine the sizes of the sets. We will see many
examples of both approaches.
In Chapter 1 we begin by introducing the basic building blocks of the the-
ory: subsets, lists and permutations, multisets, binomial coefficients, and so
on. In Section 1.2 the use of these objects is illustrated by analyzing various
applications and examples.
In Chapter 2 we introduce the idea of generating series. This begins with
the Binomial Theorem and Binomial Series, which are of fundamental im-
portance for later results. The general theory of generating series is devel-
oped in Section 2.2, and its use is illustrated by analyzing “compositions”
in Section 2.3.
In Chapter 3 we consider the enumeration of various sets of binary strings,
namely those which can be described by regular expressions – the “rational
languages”. This provides an interesting and varied class of examples to
which the results of Chapters 2 and 4 apply.
In Chapter 4 we consider sequences which satisfy a homogeneous linear
recurrence relation with initial conditions, the sequences arising in Chapters
13
2 and 3 being examples. This technique allows us to calculate the numbers
which answer the various counting problems we have been considering.
By using Partial Fractions we can derive an even better solution to such
problems, although the calculations involved are also more complicated.
(We include a proof of Partial Fractions for completeness.) In Section 4.4 we
briefly discuss recurrence relations which are quadratic rather than linear.
Two additional topics are discussed in Chapters 11 and 12.
Chapter 1
Basic Principles of Enumeration.
1.1 The Essential Ideas.
1.1.1 Choices – “AND” versus “OR”.
In the next few pages we will often be constructing an object of some kind
by repeatedly making a sequence of choices. In order to count the total
number of objects we could construct we must know how many choices are
available at each step, but we must know more: we also need to know how to
combine these numbers correctly. A generally good guideline is to look for
the words “AND” and “OR” in the description of the sequence of choices
available. Here are a few simple examples.
Example 1.1. On a table before you are 7 apples, 8 oranges, and 5 ba-
nanas.
• Choose an apple and a banana.
There are 7 choices for an apple AND 5 choices for a banana: 7×5 =
35 choices in all.
• Choose an apple or an orange.
There are 7 choices for an apple OR 8 choices for an orange: 7+8 =
15 choices in all.
• Choose an apple and either an orange or a banana.
There are 7× (8 + 5) = 91 possible choices.
15
16 Basic Principles. Section 1.1
• Choose either an apple and an orange, or a banana.
There are (7× 8) + 5 = 61 possible choices.
Generally, “AND” corresponds to multiplication and “OR” corresponds
to addition. The last two of the above examples show that it is important
to determine exactly how the words “AND” and “OR” combine in the de-
scription of the problem.
From a mathematical point of view, “AND” corresponds to the Cartesian
product of sets. If you choose one element of the set A AND you choose one
element of the set B, then this is equivalent to choosing one element of the
Cartesian product of A and B:
A×B = {(a, b) : a ∈ A and b ∈ B},
which is the set of all ordered pairs of elements (a, b) with a ∈ A and b ∈ B.
In general, the cardinalities of these sets are related by the formula
|A×B| = |A| · |B|.
Similarly, from a mathematical point of view, “OR” corresponds to the
union of sets. If you choose one element of the set A OR you choose one
element of the set B, then this is equivalent to choosing one element of the
union of A and B:
A ∪B = {c : c ∈ A or c ∈ B},
which is the set of all elements c which are either in A or in B.
It is not always true that |A∪B| = |A|+ |B|, because any elements in both
A and B would be counted twice by |A| + |B|. The intersection of A and B
is the set
A ∩B = {c : c ∈ A and c ∈ B},
which is the set of all elements c which are both in A and in B. What is
generally true is that
|A ∪B| = |A|+ |B| − |A ∩B|.
(This is the first instance of the Principle of Inclusion/Exclusion, which will
be discussed in general in Subsection 1.1.6.) In particular, if A ∩ B = ∅
then |A ∪ B| = |A| + |B|. Thus, in order to interpret “OR” as addition, it
Section 1.1 Basic Principles. 17
is important to check that the sets of choices A and B have no elements in
common. Such a union of sets A and B for which A ∩ B = ∅ is called a
disjoint union of sets.
When solving enumeration problems it is usually very useful to describe
a choice sequence for constructing the set of objects of interest, paying close
attention to the words “AND” and “OR”.
1.1.2 Lists, permutations, and subsets.
A list of a set S is a list of the elements of S exactly once each, in some
order. For example, the lists of the set {1, a,X, g} are:
1aXg a1Xg X1ag g1aX
1agX a1gX X1ga g1Xa
1Xag aX1g Xa1g ga1X
1Xga aXg1 Xag1 gaX1
1gaX ag1X Xg1a gX1a
1gXa agX1 Xga1 gXa1
A permutation is a list of the set {1, 2, ..., n} for some n ∈ N. A per-
mutation σ : a1a2...an can be interpreted as a function σ : {1, 2, ..., n} →
{1, 2, ..., n} by putting σ(i) = ai for all 1 ≤ i ≤ n.
To construct a list of S we can choose any element v of S to be the first
element in the list and follow this with any list of the set S r {v}. That is
how the table above is arranged – each of the four columns corresponds
to one choice of an element of {1, a,X, g} to be the first element of the list.
Within each column, all the lists of the remaining elements appear after the
first element.
Let pn denote the number of lists of an n-element set S. The first sentence
of the previous paragraph is translated into the equation
pn = n · pn−1,
provided that n is positive. (In this equation there are n choices for the first
element v of the list, AND pn−1 choices for the list of S r {v} which follows
it.) It is important to note here that each list of S will be produced exactly
once by this construction.
18 Basic Principles. Section 1.1
Since it is easy to see that p1 = 1 (and p2 = 2), a simple proof by induction
on n shows the following:
Theorem 1.2. For every n ≥ 1, the number of lists of an n-element set S is
n(n− 1)(n− 2) · · · 3 · 2 · 1.
In particular, taking S = {1, 2, ..., n}, this is the number of permutations of
size n. The term n factorial is used for the number n(n− 1) · · · 3 · 2 · 1, and it
is denoted by n! for convenience. We also define 0! to be the number of lists
of the 0-element (empty) set ∅. Since we want the equation pn = n · pn−1 to
hold when n = 1, and since p1 = 1! = 1, we conclude that 0! = p0 = 1 as
well.
A subset of a set S is a collection of some (perhaps none or all) of the
elements of S, at most once each and in no particular order.
To specify a particular subset A of S, one has to decide for each element v
of S whether v is in A or v is not in A. Thus we have two choices – v ∈ A OR
v 6∈ A – for each element v of S. If S = {v1, v2, ..., vn} has n elements then the
total number of choices is 2n since we have 2 choices for v1 AND 2 choices
for v2 AND . . . AND 2 choices for vn.
Theorem 1.3. For every n ≥ 0, the number of subsets of an n-element set
is 2n.
A partial list of a set S is a list of a subset of S. That is, it is a list of some
(perhaps none or all) of the elements of S, at most once each and listed in
some particular order. We are going to count partial lists of length k of an
n-element set.
First think about the particular case n = 6 and k = 3, and the set S =
{a, b, c, d, e, f}. A partial list of S of length 3 is a list xyz of elements of S,
which must all be different. There are:
6 choices for x (since x is in S), AND
5 choices for y (since y ∈ S but y 6= x), AND
4 choices for z (since z ∈ S but z 6= x and z 6= y).
Altogether there are 6 · 5 · 4 = 120 partial lists of {a, b, c, d, e, f} of length 3.
Section 1.1 Basic Principles. 19
This kind of reasoning works just as well in the general case. If S is an
n-element set and we want to construct a partial list v1v2...vk of elements of
S of length k, then there are:
n choices for v1, AND
n− 1 choices for v2, AND
....
n− (k − 2) choices for vk−1, AND
n− (k − 1) choices for vk.
This proves the following result.
Theorem 1.4. For n, k ≥ 0, the number of partial lists of length k of an
n-element set is n(n− 1) · · · (n− k + 2)(n− k + 1).
Notice that if k > n then the number 0 will appear as one of the factors in
the product n(n− 1) · · · (n− k + 2)(n− k + 1). This makes sense, because if
k > n then there are no partial lists of length k of an n-element set. On the
other hand, if 0 ≤ k ≤ n then we could also write this product as
n(n− 1) · · · (n− k + 2)(n− k + 1) = n!
(n− k)! .
We next count subsets of an n-element set S which have a particular size
k. So for n, k ≥ 0 let (n
k
)
denote the number of k-element subsets of an n-
element set S. Notice that if k < 0 or k > n then
(
n
k
)
= 0 because in these
cases it is impossible for S to have a k-element subset. Thus we need only
consider k in the range 0 ≤ k ≤ n.
To count k-element subsets of S we consider another way of constructing
a partial list of length k of S. Specifically, we can choose a k-element subset
A of S AND a list of A. The result will be a list of a subset of S of length
k. Since every partial list of length k of S is constructed exactly once in this
way, this translates into the equation(
n
k
)
· k! = n!
(n− k)! .
20 Basic Principles. Section 1.1
In summary, we have proved the following result.
Theorem 1.5. For 0 ≤ k ≤ n, the number of k-element subsets of an n-
element set is (
n
k
)
=
n!
k!(n− k)! .
The numbers
(
n
k
)
are read as “n choose k” and are called binomial coefficients .
1.1.3 Think of what the numbers mean.
Usually, when faced with a formula to prove, one’s first thought is to prove
it by algebraic calculations, or perhaps with an induction argument, or maybe
with a combination of the two. But often that is not the easiest way, nor is
it the most informative. A much better strategy is one which gives some in-
sight into the meaning of all of the parts of the formula. If we can interpret
all the numbers as counting things, addition as “OR”, and multiplication
as “AND”, then we can hope to find an explanation of the formula by con-
structing some objects in the correct way.
Example 1.6. Consider the equation, for any n ≥ 0:(
n
0
)
+
(
n
1
)
+
(
n
2
)
+ · · ·+
(
n
n
)
= 2n.
This could be proved by induction on n, but many more details would
have to be given and the proof would not address the true “meaning” of
the formula. Instead, let’s interpret everything combinatorially:
• the number of subsets of the n-element set {1, 2, ..., n} is 2n;
• for each 0 ≤ k ≤ n, the number of k-element subsets of {1, 2, ..., n}
is
(
n
k
)
;
• addition corresponds to “OR” (that is, disjoint union of sets).
So, this formula is saying that choosing a subset of {1, 2, ..., n} (in one of
2n ways) is equivalent to choosing a k-element subset of {1, 2, ..., n} (in
one of
(
n
k
)
ways) for exactly one value of k in the range 0 ≤ k ≤ n. Said
that way the formula becomes self–evident, and there is nothing more
to prove.
Section 1.1 Basic Principles. 21
Example 1.7. Consider the equation(
n
k
)
=
(
n− 1
k − 1
)
+
(
n− 1
k
)
where we are using the fact that
(
m
j
)
= 0 if j < 0 or j > m.
This equation can be proven algebraically from the formula of Theo-
rem 1.5, and that is a good exercise which I encourage you to try. But
a more informative proof interprets these numbers combinatorially as
follows:
• (n
k
)
is the number of k-element subsets of {1, 2, ..., n};
• (n−1
k−1
)
is the number of (k − 1)-element subsets of {1, 2, ..., n− 1};
• (n−1
k
)
is the number of k-element subsets of {1, 2, ..., n− 1};
• addition corresponds to disjoint union of sets.
So, this equation is saying that choosing a k-element subsetA of {1, 2, ..., n}
is equivalent to either choosing a (k−1)-element subset of {1, 2, ..., n−1}
or a k-element subset of {1, 2, ..., n − 1}. This is perhaps not as clear as
the previous example, but the two cases depend upon whether the cho-
sen k-element subset A of {1, 2, ..., n} is such that n ∈ A OR n 6∈ A. If
n ∈ A then Ar {n} is a (k− 1)-element subset of {1, 2, ..., n− 1}, while if
n 6∈ A then A is a k-element subset of {1, 2, ..., n − 1}. This construction
explains the correspondence, proving the formula.
This principle – interpreting equations combinatorially and proving the
formulas by describing explicit correspondences between sets of objects –
is one of the most important and powerful ideas in enumeration. We will
apply this way of thinking throughout the first part of these notes.
Incidentally, the equation in Example 1.7 is a very useful recurrence rela-
tion for computing binomial coefficients quickly. Together with the facts
(
n
k
)
=
n!
k!(n− k)! =
n!
(n− k)!(n− (n− k))! =
(
n
n− k
)
and
(
n
0
)
=
(
n
n
)
= 1 it can be used to compute any number of binomial coeffi-
22 Basic Principles. Section 1.1
cients without difficulty. The resulting table is known as Pascal’s Triangle :
n\k 0 1 2 3 4 5 6 7 8
0 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
7 1 7 21 35 35 21 7 1
8 1 8 28 56 70 56 28 8 1
1.1.4 Multisets.
Imagine a bag which contains a large number of marbles of three colours
– red, green, and blue, say. The marbles are all indistinguishable from one
another except for their colours. There are N marbles of each colour, where
N is very, very large (more precisely we should be considering the limit
as N → ∞). If I reach into the bag and pull out a handful of 11 marbles,
I will have r red marbles, g green marbles, and b blue marbles, for some
nonnegative integers (r, g, b) such that r + g + b = 11. How many possibile
outcomes are there?
The word “multiset” is meant to suggest a set in which the objects can oc-
cur more than once. For example, the outcome (4, 5, 2) in the above situation
corresponds to the “set” {R,R,R,R,G,G,G,G,G,B,B} in which R is a red
marble, G is a green marble, and B is a blue marble. This is an 11-element
multiset with elements of three types. The number of these multisets is the
solution to the above problem.
Definition 1.8. Let n ≥ 0 and t ≥ 1 be integers. A multiset of size nwith
elements of t types is a sequence of nonnegative integers (m1, ...,mt)
such that
m1 +m2 + · · ·+mt = n.
The interpretation is thatmi is the number of elements of the multiset which
are of the i-th type, for each 1 ≤ i ≤ t.
Section 1.1 Basic Principles. 23
Theorem 1.9. For any n ≥ 0 and t ≥ 1, the number of n-element multisets
with elements of t types is (
n+ t− 1
t− 1
)
.
Proof. Think of what that number means! By Theorem 1.5,
(
n+t−1
t−1
)
is the
number of (t − 1)-element subsets of an (n + t − 1)-element set. So, let’s
write down a row of (n+ t− 1) circles from left to right:
O O O O O O O O O O O O O
and cross out some t− 1 of these circles to choose a (t− 1)-element subset:
O O O O X O O O O O X O O
Now the t − 1 crosses chop the remaining sequence of n circles into t seg-
ments of consecutive circles. (Some of these segments might be empty,
which is to say of length zero.) Let mi be the length of the i-th segment
of consecutive O-s in this construction. Then m1 + m2 + · · · + mt = n, so
that (m1,m2, ...,mt) is an n-element multiset with t types. Conversely, if
(m1,m2, ...,mt) is an n-element multiset with t types then write down a se-
quence of m1 O-s, then an X, then m2 O-s, then an X, and so on, finishing
with an X and then mt O-s. The positions of the X-s will indicate a (t − 1)-
element subset of the positions {1, 2, ..., n+ t− 1}.
The construction of the above paragraph shows how to translate between
(t − 1)-element subsets of {1, 2, ..., n + t − 1} and n-element multisets with
t types of element. This one–to–one correspondence completes the proof of
the theorem.
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MATH 239 QF5203 INFR08010 IT114105 FIT2093 ECON3124 ECON1195 MATH 337 STAT 3006 COMP202 COMP3130 MATH3609 COMP 250 CS6140 IEEE ELE00041I FIN40090 CIV 5899 MATH3066 IGNMENT MATH 2009 STAT3405 HIST 101 MATH 3281 CEGE0042 ICT532 LAB 7 LAB 5 MCO455 CSIT214 Q4 IFB104 A1 ECO220Y5Y DA5020 CMPSC473 FIN9007 DSCI553 MSCI534 COMP3308 CSE-112 CMPT 310 EE104 F 2C P20 ALY2010 BUSM060 STA258 ECO00030M R BIOSTAT 274 COS6012-B FIT3155 CMPUT 340 EC421 INFS4205 ECOM135 NATS 1740 COMP809 MBE 6005 STA465 CSSE2310 CIVL2700 DATA4700 CMPSC 473 ISE 529 CS5044 MAST30020 COMP2207 ENGG1001 COMP0046 FIT3165 STAT GR5241 MECH0007 vATHK1001 ANALYTIC ATHK1001 MATH352001 HW3 S1 BMAN 24662 COMP 636 Stat-461 STAT 302 STAT 2132 STAR534 ENG4137 CMP3752M CS3103 BISM3222 STA 135 ME155A IIMT3601 ECON3371 ECON 3371 HW4 FINA 5840 DT211C CTEC3902 MAT005 MATH2148 S261F Econ 332 MATH3734 MIS41270 7CCMFM13 SYST 664 CIS434 CENG10014 MT4111 MT4527 ECMM424 CMT219 Math 136 MATH 3120 COMP10002 MAST20009 MATH0058 BEEM117 ISYE 6420 MGMTMFE 405 25579 1ECE 273 M122 COMP4434 1ELEC ELEC5620M STAT0010 BFF2401 MXB361 MATH3811 CS330 CMT307 FIT1033 CptS 583 UP 431 MATH6005 AF5343 CS2435 ECOM30002 ES50061 CS 2506 ECMT6002 CODE 00099F MT4539 ESSAY ICM317 MATH1231 MFIT5010 BMA547 MATH334 CIVL2201 MATH39032 T-04 CIVE97119 EMESTER CAB302 CAB301 59PM STAT802 ISYS3412 RM3 COSC 2123 AcF 311 MATH2022 1FINAL 4CCS1PHC ICM 142 ENVS3023 Math 4023 I1 ACSE8 FIN B385F CS-275 STA 4109-01 A31021 EE140 MT1 EE124 ECO-6004B Inf2-IADS CS126 IS2184 UL18 MATH5092 ECOM009 COMP 6211 ELEC202 MKT3019 IAB230 FIT5046 584-S21 COMP3210 CO250 A29401 MATH 136 MGT6174 ECMT6007 DATA2001 MAT 443 MATH10222 PHDE0071 ELE00123M COSC1295 MAT00045H MATH32032 1004GRC MAS223 COMP10001 DATA3404 CIVL2110 STAT2008 FNCE10002 EEE6431 COM6515 COMP9120 NBS8185 ECTE942 ECON 405 MATH314301 STAT 2011 COMPSCI 350 COMP20007 CMPSC-132 ECON5002 STA 35 MTH6108 FIT5147 MTH6150 100C FINA 5260 DESC9115 ECON4998 NBS8257 GMM ECON 6070 BSA2 ECE599 STATS 102B STA120 INFS601 BFC3241 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COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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