Table Of ContentCombinatorial Mathematics Notes
UIUC MATH 580, F’08
Jingjin Yu
2
Chapter 1
Combinatorial Arguments
1.1 Classical Models
Topic: Elementary Principles
Remark. Sum Principle: If a finite set A is partitioned into sets B1,...,Bk, then A
k
i 1 Bi . Product Principle: If the elements of set A are built via successive choices, w(cid:0) h(cid:0)er(cid:2)e
the number of options for the ith choice is independent of the outcomes of the earlier choices,
(cid:0) (cid:0) (cid:0)
t(cid:0)hen A is the product of the number of options for the successive choices.
Rem(cid:0)ar(cid:0)k. Principle of Counting in Two Ways: When two formulas count the same set,
their values are equal.
Remark. Bijection Principle: If there is a bijection from one set to another, then the two
sets have the same size.
Remark. Pigeonhole Principle: The maximum in a set of numbers is at least as large
as the average (and the minimum is at least as small). In particular, placing more than kn
objects into n boxes puts more than k objects into some box.
Remark. Polynomial Principle: If two polynomials in x are equal for infinitely many
values of x, then they are the same polynomial (and equal for all real x). The analogous
statement holds with more variables by induction on the number of variables.
Topic: Words, Sets and Multisets
Definition 1.1.1 A k-word or word of length k is a list of k elements from a given set (the
alphabet); we may call the elements “letters”. A simple word is a word whose letters are
distinct. A k-set is a set with k elements; a k-set in a set S is a subset of S with k elements.
n
We use , read “n choose k”, to denote the number of k-sets in an n-set. A multiset from
k
a set S is a selection from S with repetition allowed.
(cid:2) (cid:3)
3
k
Proposition 1.1.2 Therearen wordsoflengthk fromanalphabetS ofsizen. Equivalently,
k
there are n functions from k to S.
(cid:3) (cid:4)
Definition 1.1.3 For n 0, we define the following products:
(cid:0)
n(cid:5) 1
n factorial: n! n i .
(cid:2)
i 0
(cid:2) (cid:6) (cid:7) (cid:8)
(cid:4)
(cid:0) k 1
falling factorial: n k (cid:2) n i .
i 0
(cid:3) (cid:4) (cid:2) (cid:6) (cid:7) (cid:8)
(cid:4)
k(cid:0)1
k
rising factorial: n n i .
(cid:2)
(cid:3) (cid:4) i 0
(cid:2) (cid:6) (cid:9) (cid:8)
(cid:4)
(cid:0)
Proposition 1.1.4 The number of simple k-words from an alphabet of size n is n k .
(cid:3) (cid:4)
Proposition 1.1.5 The number of k-element multisets from n , or equivalently the number
n n k 1 n k 1
of solutions to i 1xi k in nonnegative integers is n 1 (cid:3) (cid:4) k .
(cid:5) (cid:2) (cid:5) (cid:2)
(cid:0) (cid:2) (cid:2) (cid:2)
Definition 1.1(cid:0).6 A composition of the positive int(cid:2)eger k(cid:3)is a(cid:2)list of(cid:3)positive integers sum-
ming to k. The entries in the list are the parts of the compositions.
k 1
Corollary 1.1.7 There are compositions of k with n parts.
n 1
(cid:2)
(cid:2)
Exercise 1.1.2. For each di(cid:2)ce th(cid:3)e probability of getting even/odd is 1 2. Then for first two
dices, the probability of getting a total of even is again 1 4 1 4 1 2. By induction, the
(cid:10)
probability is 1 2.
(cid:10) (cid:9) (cid:10) (cid:2) (cid:10)
Exercise 1.1.3(cid:10). For n, we need to count the possible distinct consecutive segments of length
from 1 to n. There are n n 1 n 2 ... 1 of these. We do the same for m and
multiply them to get the number.
(cid:9)(cid:6) (cid:7) (cid:8)(cid:9)(cid:6) (cid:7) (cid:8)(cid:9) (cid:9)
30 4 1
Exercise 1.1.5. Firstproblemistopickfourseparatorsfrom33objects: . For second
4 1
one, we may assume that one candidate gets more than 15 votes. Then th(cid:5)e(cid:2)rest splits 1 to
14 votes; there are 14 i 3 1 of outcomes. Subtracting these for each(cid:2)ca(cid:2)ndid(cid:3)ate from the
i 0 3 1
(cid:5) (cid:2) 30 4 1 14 i 3 1
total, we get the possible outcomes as 4 .
(cid:0) (cid:2) 4 1 i 0 3 1
(cid:0) (cid:2) (cid:3) (cid:5) (cid:2) (cid:5) (cid:2)
Exercise 1.1.7. a)Allthesubsets,minus(cid:2)theon(cid:7)eswit(cid:0)h only(cid:2)even numbers... b)Thek choices
(cid:2) (cid:3) (cid:0) (cid:2) (cid:3)
need at least 2k 1 elements. When n 2k 1, there is only a single choice. The choices are
n k 1
then k . c)(cid:7)The first one is just th(cid:2)e prod(cid:7)uct rule: one choice for An n , then n choices
for An (cid:2)1 (cid:5)etc. the answer is then n!. For the second question, we may first pick k elements
(cid:2)(cid:3) (cid:4)
from(cid:2)n(cid:2) and(cid:3) assume An is these k elements. We then permute these elements and insert n
bars to get An 1 to A0. Summing up these should give us the answer.
(cid:3) (cid:4)
Exercise 1.1.(cid:2)10. We first pick 1 number from 1 9 for the first digit. We then have two
cases: 1. The chosen number is used again 2. The chosen number is not used again. We have
(cid:7)
4
4 5
9 numbers left, and there are either 9 or 9 ways of constructing the rest. We exclude from
these the cases where some number is used 3 or 4 times.
Exercise 1.1.14. This is simply counting all possible compositions for k 2.
Exercise 1.1.16. (cid:10)
Exercise 1.1.18. May treat the two disjointly; the runs separate m into k 1 pieces; it also
puts k bars in n, with no two bars are consecutive.
(cid:7)
n
Exercise 1.1.21. Every four points gives a pair of crossing chords, .
4
(cid:2) (cid:3)
1.2 Identities
Topic: Lattice Paths and Pascal’s Triangle
Definition 1.2.1 Alattice path isalattice walk inwhicheach stepincreases onecoordinate.
n
Remark. For lattice path of length n with height k, there are of these. We also obtain
k
Pascal’s Formula by noticing that there are two ways in the last step, one horizontal and
one vertical. Therefore, n n 1 n 1 . (cid:2) (cid:3)
k k k 1
(cid:2) (cid:2)
(cid:2) (cid:9) (cid:2)
Theorem 1.2.2 (1.2.3(cid:2)E(cid:3)lem(cid:2)enta(cid:3)ry (cid:2)Iden(cid:3)tities) a
n n
1
k n k
(cid:8) (cid:5)n(cid:6)(cid:2)(cid:5)n 1(cid:6) n 1
2 (cid:7)
k k k 1
(cid:7) (cid:7)
(cid:8) (cid:5) n(cid:6)(cid:2)(cid:5) n (cid:6)1(cid:9)(cid:5) (cid:6) k n n n l
3 k n (cid:7)
k k 1 l k l k l
(cid:7) (cid:7)
(cid:8) (cid:5) (cid:6)n (cid:2) (cid:5) (cid:6) (cid:5) (cid:6)(cid:5)n(cid:6) (cid:2) (cid:5) (cid:6)(cid:5) (cid:6)
n k n
4 2 (cid:7) r r 1 (cid:7)
k k
k k
(cid:8) n (cid:5) (cid:6)(cid:2) n (cid:5) (cid:6)(cid:2) (cid:6) (cid:9) (cid:8)
(cid:7) k n 1 (cid:7) m k n k m n 1
5
r r 1 r s r s 1
k 0 (cid:9) k m (cid:9) (cid:7) (cid:9) (cid:9)
(cid:8) (cid:5) (cid:6)2(cid:2)(cid:5) (cid:6) (cid:5) (cid:6)(cid:5) (cid:6) (cid:2)(cid:5) (cid:6)
(cid:7)(cid:0) n 2n(cid:9) (cid:0)(cid:7)(cid:2) m n m n (cid:9) (cid:9)
6
k n k r k r
k k (cid:9)
(cid:8) (cid:5) (cid:6) (cid:2)(cid:5) (cid:6) (cid:5) (cid:6)(cid:5) (cid:6) (cid:2)(cid:5) (cid:6)
(cid:7) (cid:7) (cid:7)
u k u k 1
Remark. Extended binomial coefficient 1 .
k k
(cid:2) (cid:5) (cid:2)
(cid:2) (cid:6)(cid:7) (cid:8)
Theorem 1.2.3 (1.2.6 Extended Binomia(cid:2)l T(cid:3)heorem)(cid:2)For u(cid:3),x with x 1, 1
u u k (cid:2)
x x .
k 0 k (cid:5) (cid:0) (cid:0) (cid:11) (cid:6) (cid:9)
(cid:8) (cid:2) (cid:6)
Topic:(cid:0)Dela(cid:2)n(cid:3)noy Numbers
5
Definition 1.2.4 The Delannoy number dm,n is the number of paths from 0,0 to m,n
such that each move is by one of 1,0 , 0,1 , 1,1 . The number of the form dn,n are the
(cid:6) (cid:8) (cid:6) (cid:8)
central Delannoy numbers.
(cid:12)(cid:6) (cid:8) (cid:6) (cid:8) (cid:6) (cid:8)(cid:13)
m n k
Proposition 1.2.5 (1.2.10) The Delannoy number dm,n k k m .
(cid:5)
(cid:2)
Definition 1.2.6 The Hamming ball of radius m in n dim(cid:0)en(cid:2)sio(cid:3)n(cid:2)s is(cid:3)the set consisting of
n
the lattice points in that are within m unit coordinate steps from one point (such as the
(cid:3)
origin).
n n m k
Proposition 1.2.7 The size of the Hamming ball of radius m in is 2 .
k k k
(cid:3)
Proof. k denotes the number of non-zero coordinates ( nk ); their abs(cid:0)olut(cid:2)e(cid:3)v(cid:2)alu(cid:3)e sum to at
m
most m, this is equivalent to k 1 positive integers summing to m 1 ( ). The coordinates
k
can be positive or negative (2k). (cid:2) (cid:3) (cid:2)
(cid:9) (cid:9)
(cid:2) (cid:3)
m n k n m k
Theorem 1.2.8 (1.2.13) For m,n 0, k k m k k k 2 .
(cid:0) (cid:5)
(cid:5) (cid:2)
Exercise 1.2.2. Both are just choosing k(cid:0)from(cid:2) m(cid:3)(cid:2) n(cid:3). (cid:0) (cid:2) (cid:3)(cid:2) (cid:3)
m k 1 m k 1
Exercise 1.2.4. k m 1 ; then apply(cid:9)ing Pascal’s formula to get the sum. The
other side is the same(cid:5). (cid:2) (cid:5) (cid:2)
(cid:2) (cid:2)
(cid:2) (cid:3) (cid:2) (cid:3)
Exercise 1.2.6. Follow the formula...
k
Exercise 1.2.7. 1 .
k
Exercise 1.2.8. (cid:6)C(cid:7)ou(cid:8)nting i by counting in two ways: one is the k-words from i , the other
2
is number of repeated elements used. For example for i , if we use all different items, there
(cid:3) (cid:4)
i i
are different ways of picking two elements; the ordering matters so 2 total. Then there
2 2
i
are different ways of picking one element and use it twice (no order here). Therefore,
1
2 (cid:2) (cid:3)i i (cid:2) (cid:3)
i 2 .
2 1
(cid:2) (cid:3)
Ex(cid:2)ercise(cid:9)1.2.10. If two marbles are put in the same jar, it corresponds to a diagonal step;
(cid:2) (cid:3) (cid:2) (cid:3)
so it is easy to establish a bijection.
1.3 Applications
Topic: Graphs and Trees
n
Corollary 1.3.1 There are 2 2 trees with vertex set n .
(cid:6) (cid:8)
(cid:3) (cid:4)
Definition 1.3.2 The functional digraph of a function f : S S is the directed graph
having vertex set S and an edge from x to f x for each x S. There are S edges, and each
(cid:14)
vertex is the tail exactly once.
(cid:6) (cid:8) (cid:5) (cid:0) (cid:0)
6
n 2
Theorem 1.3.3 There are n trees with vertex set n .
(cid:2)
(cid:3) (cid:4)
Topic: Multinomial Coefficients
n
Proposition 1.3.4 (1.3.5) Thereare m! i 1ki!wordsof length n havingexactly ki letters
n
of type i, where i 1ki m. (cid:10) (cid:0)
(cid:8)
(cid:0) (cid:2)
Corollary 1.3.5(cid:0)(1.3.6) The number of trees with vertex set n in which vertices 1,...,n
n 2 !
have degrees d1,...,dn, respectively, is d 1 !. (cid:3) (cid:4)
i
(cid:3) (cid:2) (cid:4)
(cid:0)
(cid:3) (cid:2) (cid:4)
n n
k k
k
Proposition 1.3.6 (1.3.8 Multinomial Theorem) Forn 0, xi xii.
(cid:0) k1,...,kn
i 1 i 1
(cid:5) (cid:2) (cid:5) (cid:6)
(cid:9)(cid:7) (cid:10) (cid:7) (cid:4)
(cid:0) (cid:0)
Topic: The Ballot Problem
Theorem 1.3.7 (1.3.12 Bertrand) Among the lists formed from a copies of A and b copies
a b a b
of B, there are such that every initial segment has at least as many As as Bs.
a a 1
(cid:5) (cid:5)
(cid:7) (cid:5)
Lemma 1.3.8 (cid:2)(1.3(cid:3).13)(cid:2) Th(cid:3)e central binomial coefficient 2n counts the following types of
n
lattice paths of length 2n that start at 0,0
(cid:2) (cid:3)
(cid:6) (cid:8)
A) Those ending at n,n .
(cid:6) (cid:8)
B) Those never rising above the line y x.
(cid:2)
C) Those never returning to the line y x.
(cid:2) 2k 2n 2k n
Theorem 1.3.9 (1.3.14) For n 0, k 0n k n k 4 .
(cid:0) (cid:2)
(cid:5) (cid:0) (cid:2) (cid:2)
Topic: Catalan Numbers (cid:0) (cid:2) (cid:3)(cid:2) (cid:3)
1 2n
Definition 1.3.10 The Catalan sequence is defined by Cn n 1 n for n 0. The
number Cn is called the nth Catalan number.
(cid:2) (cid:5) (cid:15)
(cid:2) (cid:3)
Theorem 1.3.11 (1.3.16) If p,q are relatively prime positive integers, then the number of
1 p q
lattice paths from 0,0 to p,q that do not rise above the line py qx is .
p q p
(cid:5)
(cid:6) (cid:8) (cid:6) (cid:8) (cid:2) (cid:5)
Theorem 1.3.12 (1.3.21) The number of binary trees with n 1 leaves is th(cid:2)e nt(cid:3)h Catalan
1 2n
number, .
n 1 n (cid:9)
(cid:5)
Example 1.(cid:2)3.1(cid:3)3 (1.3.22) There are 1 2n triangulations of a convex n 2 -gon.
n 1 n
(cid:5) (cid:6) (cid:9) (cid:8)
(cid:2) (cid:3)7
Exercise 1.3.2. n iki. We may then first count the number of ways to partition n into
m blocks of size iki (simply multinomial formula). Then for each iki, it is further partitioned
(cid:2)
into ki blocks of size i(cid:0)(multinomial formula again). Multiplying gives us the result.
Exercise 1.3.3. The first row decides the second row; therefore we only need to work with
the first row. We establish a bijection this way: suppose the numbers of the first row are
a1,...,an. Let k ai 1 ai 1, we construct the lattice path by add one horizontal step
followed by k vertical ones. This gives us a bijection (needs to add extra stuff).
(cid:2) (cid:5) (cid:7) (cid:7)
Exercise 1.3.4.Catalan number over n factorial.
Exercise 1.3.5. Probability that the lattice path never goes above y x 1 after first step.
a b 1 a b 1 a b
Total paths this way is . The denominator is again .
a 1 a (cid:2)a (cid:9)
(cid:5) (cid:2) (cid:5) (cid:2) (cid:5)
n 1 2n
Exercise 1.3.6. 4 2 (cid:2) : t(cid:7)he total possible paths minus these that one never trails.
(cid:2)n n (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
(cid:7)
(cid:2) (cid:3)
8
Chapter 2
Recurrence Relations
2.1 Obtaining Recurrences
Topic: Classical Examples
Example 2.1.1 Regions in theplanebynlines suchthat nothreelines have acommon point:
an an 1 n.
(cid:2) (cid:2) (cid:9)
Example 2.1.2 Fibonacci numbers
Example 2.1.3 Derangements of n objects.
n n
Dn n! k 1 k Dn k for n 1, with D0 1.
(cid:2) (cid:7) (cid:0) (cid:2) (cid:15) (cid:2)
Dn n (cid:0)1 Dn(cid:2) 1(cid:3) Dn 2 for n 2, with D0 1 and D1 0.
(cid:2)(cid:6) (cid:7) (cid:8)(cid:6) (cid:2) (cid:9) (cid:2) (cid:8) (cid:15)n (cid:2) (cid:2)
Example 2.1.4 Catalan number Cn k 1Ck 1cn k, with C0 1.
(cid:2) (cid:0) (cid:2) (cid:2) (cid:2)
Topic: Variations (cid:0)
Example 2.1.5 Arrangements with n distinct available objects.
C(n, k) = C(n-1, k) + C(n-1, k-1) #k-subsets of n .
(cid:3) (cid:4)
P(n, k) = P(n-1,k) + kP(n-1, k-1) # simple k-words from n .
(cid:3) (cid:4)
S(n, k) = kS(n-1,k) + S(n-1, k-1) # partitions of n with k blocks.
(cid:3) (cid:4)
c(n, k) = (n-1)c(n-1, k) + c(n-1, k-1) # permutations of n with k cycles.
(cid:3) (cid:4)
9
Proposition 2.1.6 (2.1.10) Thenumber of lattice points within m lattice steps of the origin
inn-dimensionalspacesatisfies therecurrenceam,n am,n 1 am 1,n 1 am 1,n form,n 0.
(cid:2) (cid:2) (cid:9) (cid:2) (cid:2) (cid:9) (cid:2) (cid:16)
Exercise 2.1.2. an 1 gives us n 1 pairs. Addingp2n 1,p2n as a pair gives a pairing (an 1 of
these). Then we may pick a pair from the n 1 pairs and pair them with the last two persons
(cid:2) (cid:7) (cid:2) (cid:2)
(2 n 1 an 1 of these). Total is then 2n 1 an 1.
(cid:7)
2n n n 2n n 2n n 1
Ex(cid:6)er(cid:7)cis(cid:8)e 2(cid:2).1.5. an,k 2k k (cid:6) k(cid:7) (cid:8) 2(cid:2)k an,k, so k 1 2k 2 an,k 1, k 1
2n 2 2n 2n 2n 2 (cid:2)
2k 2 an 1,k 1. Theref(cid:2)ore 2k(cid:10) an,k(cid:17) 2k (cid:2)2 an,k(cid:10)1 2k 2 (cid:2)an 1(cid:2),k 1. (cid:2) (cid:10) (cid:2) (cid:2) (cid:2)
(cid:2) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2)(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
Ex(cid:2)erc(cid:10)ise(cid:2)2.1(cid:2).6. Fn 1. (cid:10) (cid:2) (cid:2) (cid:10) (cid:2) (cid:9) (cid:2) (cid:10) (cid:2) (cid:2)
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
Exercise 2.1.7. an(cid:5) 2an 1 an 3. It would be 2an 1 but we need to remove the of an 1
that ends with 01. Th(cid:2)ese ar(cid:2)e j(cid:7)ust a(cid:2)n 3 plus 01 at the e(cid:2)nd. (cid:2)
(cid:2)
2.2 Elementary Solution Methods
Topic: The Characteristic Equation Method
2
Example 2.2.1 Fn Fn 1 Fn 2 x x 1 0. Solving this gives two roots a1,a2.
n n
Then Fn c1a1 c2a(cid:2)2. P(cid:2)lug(cid:9)ging(cid:2)in t(cid:17)he init(cid:7)ial c(cid:7)ondi(cid:2)tion gives us the constants.
(cid:2) (cid:9) k n
Proposition 2.2.2 Considerarecurrencean i 1cian i f n suchthatf n F n c ,
where F is a polynomial of degree d. If c has multiplicity r as a characteristic root of the
(cid:2)(cid:6) (cid:0) (cid:2) (cid:8)(cid:9) (cid:6) (cid:8) (cid:6) (cid:8) (cid:2) (cid:6) (cid:8)
homogeneous part (r may be 0), then the recur(cid:0)rence has a solution of the form P n nrcn,
where P is a polynomial of degree at most d.
(cid:6) (cid:8)
Topic: The Generating Function Method
Algorithm. Thegenerating function method usesthefollowingstepstosolvearecurrence
for a .
(cid:18) (cid:19)
1) Sum the recurrence over its “region of validity” (the values of the parameter where the
recurrence holds) to introduce the generating function A x and obtain an equation that
A x satisfies.
(cid:6) (cid:8)
2) So(cid:6)lv(cid:8)e this equation to express A x in terms of x.
3) Find the formal power series exp(cid:6)an(cid:8)sion and set an xn A x .
(cid:2)(cid:3) (cid:4) (cid:6) (cid:8) k
Lemma 2.2.3 (2.2.17) For k , the power series expansion of 1 cx is
(cid:0)
(cid:2)
(cid:5) (cid:6) (cid:7) (cid:8)
1 n k 1
n n
c x .
1 cx k (cid:7) k 1
n 0 (cid:9) (cid:7)
(cid:2) (cid:5) (cid:6)
(cid:6) (cid:7) (cid:8) (cid:7)(cid:0) (cid:7)
10
