An excursion through elementary mathematics, volume III discrete mathematics and polynomial algebra

An Excursion through Elementary Mathematics, Volume III_ Discrete Mathematics and Polynomial AlgebraThis is the final volume of a series of three volumes (the other ones being 9 and 8) devoted to the mathematics of mathematical olympiads. Generally speaking, they are somewhat expanded versions of a collection of six volumes, first published in Portuguese by the Brazilian Mathematical Society in 2012 and currently in its second edition.The material collected here and in the other two volumes is based on course notes that evolved over the years since 1991, when I first began coaching students of Fortaleza to the Brazilian Mathematical Olympiad and to the International Mathematical Olympiad. Some ten years ago, preliminary versions of the Portuguese texts also served as textbooks for several editions of summer courses delivered atUFC to math teachers of the Cape Verde Republic.

Problem Books in Mathematics

Antonio Caminha Muniz Neto

An Excursion

through Elementary Mathematics,

Volume III

Discrete Mathematics and Polynomial Algebra

Problem Books in Mathematics

Antonio Caminha Muniz Neto

An Excursion through

Elementary Mathematics, Volume III

Discrete Mathematics and Polynomial Algebra

123

Universidade Federal do Ceará

Fortaleza, Ceará, Brazil

ISSN 0941-3502 ISSN 2197-8506 (electronic)

Problem Books in Mathematics

ISBN 978-3-319-77976-8 ISBN 978-3-319-77977-5 (eBook)

https://doi.org/10.1007/978-3-319-77977-5

Library of Congress Control Number: 2017933290

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E horas sem conta passo, mudo,

This is the final volume of a series of three volumes (the other ones being [9] and[8]) devoted to the mathematics of mathematical olympiads Generally speaking,they are somewhat expanded versions of a collection of six volumes, first published

in Portuguese by the Brazilian Mathematical Society in 2012 and currently in itssecond edition

The material collected here and in the other two volumes is based on coursenotes that evolved over the years since 1991, when I first began coaching students

of Fortaleza to the Brazilian Mathematical Olympiad and to the International ematical Olympiad Some ten years ago, preliminary versions of the Portuguesetexts also served as textbooks for several editions of summer courses delivered atUFC to math teachers of the Cape Verde Republic

Math-All volumes were carefully planned to be a balanced mixture of a smooth andself-contained introduction to the fascinating world of mathematical competitions,

as well as to serve as textbooks for students and instructors involved with math clubsfor gifted high school students

Upon writing the books, I have stuck myself to an invaluable advice of theeminent Hungarian-American mathematician George Pólya, who used to say that

one cannot learn mathematics without getting one’s hands dirty That’s why, in

several points throughout the text, I left to the reader the task of checking minoraspects of more general developments These appear either as small omitted details

in proofs or as subsidiary extensions of the theory In this last case, I sometimesrefer the reader to specific problems along the book, which are marked with an *and whose solutions are considered to be an essential part of the text In general,

in each section I collect a list of problems, carefully chosen in the direction ofapplying the material and ideas presented in the text Dozens of them are taken fromformer editions of mathematical competitions and range from the almost immediate

to real challenging ones Regardless of their level of difficulty, generous hints, oreven complete solutions, are provided to virtually all of them

As a quick look through the Contents pages readily shows, this time we centrate on combinatorics, number theory, and polynomials Although the chapters’

con-vii

viii Preface

names quickly link them to one of these three major themes, whenever possible ordesirable later chapters revisit or complement material covered in earlier ones Wenow describe, a bit more specifically, what is covered within each major topic.Chapters1through5are devoted to the study of basic combinatorial techniquesand structures We start by reviewing the elementary counting strategies, emphasiz-ing the construction of bijections and the use of recursive arguments throughout Wethen go through a bunch of more sophisticated tools, as the inclusion-exclusion prin-ciple and double counting, the use of equivalence relations, metrics on finite sets,and generating functions Turning our attention to the existence of configurations,the pigeonhole principle of Dirichlet and invariants associated with algorithmicproblems now play the central role Our tour through combinatorics finishes bystudying some graph theory, all the way from the basic definitions to the classicaltheorems of Euler (on Eulerian paths), Cayley (on the number of labeled trees), andTurán (on complete subgraphs of a given graph), to name just a few ones

We then turn to elementary number theory, which is the object of Chaps.6 12

We begin, of course, by introducing the basic concepts and properties concernedwith the divisibility relation and exploring the notion of greatest common divisorand prime numbers Then we turn to diophantine equations, presenting Fermat’sdescent method and solving the famous Pell’s equation Before driving through asystematic study of congruences, we make an interlude to discuss the basics ofmultiplicative arithmetic functions and the distribution of primes, these two chaptersbeing almost entirely independent of the rest of the book From this point untilChap.12, we focus on the congruence relation and its consequences, from the verybeginnings to the finite fieldZp, primitive roots, Gauss’ quadratic reciprocity law,and Fermat’s characterization of integers that can be written as the sum of twosquares All of the above material is, here more than anywhere else in the book,illustrated with lots of interesting and challenging examples and problems takenfrom several math competitions around the world

The last nine chapters are devoted to the study of complex numbers andpolynomials Apart from what is usually present in high school classes—as thebasics of complex numbers and the notion of degree, the division algorithm, andthe concept of root for polynomials—we discuss several nonstandard topics We

begin by highlighting the use of complex numbers and polynomials as tags in

certain combinatorial problems and presenting a complete proof of the fundamentaltheorem of algebra, accompanied with several applications Then, we study thefamous theorem of Newton on symmetric polynomials and the equally famousNewton’s inequalities The next theme concerns interpolation of polynomials, whenparticular attention is placed on Lagrange’s interpolation theorem Such a result

is used to solve linear systems of Vandermonde with no linear algebra, which inturn allows us to, later, analyze an important particular class of linear recurrencerelations The book continues with the study of factorization of polynomials overQ,

Z, and Zp, together with several interesting problems on irreducibility Algebraicand transcendental numbers then make their appearance; among other topics, wepresent a simple proof of the fact that the set of algebraic numbers forms a fieldand discuss the rudiments of cyclotomic polynomials and transcendental numbers

The final chapter develops the most basic aspects of complex power series, whichare then used, disguised as complex generating functions, to solve general linearrecurrence relations.

Several people and institutions contributed throughout the years for my efforts

of turning a bunch of handwritten notes into these books The State of CearáMathematical Olympiad, created by the Mathematics Department of the FederalUniversity of Ceará (UFC) back in 1980 and now in its 37th edition, has sincethen motivated hundreds of youngsters of Fortaleza to deepen their studies ofmathematics I was one such student in the late 1980s, and my involvement with thiscompetition and with the Brazilian Mathematical Olympiad a few years later had adecisive influence on my choice of career Throughout the 1990s, I had the honor

of coaching several brilliant students of Fortaleza to the Brazilian MathematicalOlympiad Some of them entered Brazilian teams to the IMO or other internationalcompetitions, and their doubts, comments, and criticisms were of great help inshaping my view on mathematical competitions In this sense, sincere thanks go

to João Luiz de A A Falcão, Roney Rodger S de Castro, Marcelo M de Oliveira,Marcondes C França Jr., Marcelo C de Souza, Eduardo C Balreira, Breno de A A.Falcão, Fabrício S Benevides, Rui F Vigelis, Daniel P Sobreira, Samuel B Feitosa,Davi Máximo A Nogueira, and Yuri G Lima

Professor João Lucas Barbosa, upon inviting me to write the textbooks to theAmílcar Cabral Educational Cooperation Project with Cape Verde Republic, hadunconsciously provided me with the motivation to complete the Portuguese version

of these books The continuous support of Professor Hilário Alencar, president ofthe Brazilian Mathematical Society when the Portuguese edition was first published,was also of great importance for me Special thanks go to my colleagues—professors Samuel B Feitosa and Fernanda E C Camargo—who read the entireEnglish version and helped me improve it in a number of ways If it weren’t for

my editor at Springer-Verlag, Mr Robinson dos Santos, I almost surely would nothave had the courage to embrace the task of translating more that 1500 pages fromPortuguese into English I acknowledge all the staff of Springer involved with thisproject in his name

Finally, and mostly, I would like to express my deepest gratitude to my parentsAntonio and Rosemary, my wife Monica, and our kids Gabriel and Isabela Fromearly childhood, my parents have always called my attention to the importance of

a solid education, having done all they could for me and my brothers to attend thebest possible schools My wife and kids fulfilled our home with the harmony andsoftness I needed to get to endure on several months of work while translating thisbook

December 2017

1 Elementary Counting Techniques 1

2 More Counting Techniques 33

3 Generating Functions 67

4 Existence of Configurations 95

5 A Glimpse on Graph Theory 125

6 Divisibility 157

7 Diophantine Equations 193

8 Arithmetic Functions 209

9 Calculus and Number Theory 221

10 The Relation of Congruence 243

11 Congruence Classes 269

12 Primitive Roots and Quadratic Residues 283

13 Complex Numbers 317

14 Polynomials 347

15 Roots of Polynomials 363

16 Relations Between Roots and Coefficients 395

17 Polynomials OverR 417

18 Interpolation of Polynomials 435

19 On the Factorisation of Polynomials 451

xi

20 Algebraic and Transcendental Numbers 477

21 Linear Recurrence Relations 505

22 Hints and Solutions 529

Glossary 635

Bibliography 637

Index 639

Chapter 1

Elementary Counting Techniques

In this first chapter, we develop the usual elementary tools for counting the number

of distinct configurations corresponding to a certain combinatorial situation, withoutneeding to list them one by one As the reader will see, the essential ideas are theconstruction of bijections and the use of recursive arguments

Although we develop all material from scratch, the reader is expected to havesome previous experience with elementary counting techniques, and is warned thatthe material collected here can be somewhat terse at places

1.1 The Bijective Principle

In all that follows, we assume that the reader has a relative acquaintance with sets

and elementary operations on them Given n ∈ N, we let I ndenote the set

I n = {j ∈ N; 1 ≤ j ≤ n}

of natural numbers from 1 to n.

A set A is finite if A = ∅ or if there exists a bijection f : I n → A, for some

n ∈ N If A = ∅ is finite and f : I n → A is a bijection, then letting a j = f (j) we write A = {a1, , a n } and say that n is the number of elements of A (Problem1shows that this is a well defined notion) Also in this case, we write

|A| = n or #A = n

to mean that A has n elements For the sake of completeness, we say that∅ has 0elements and write|∅| = 0

© Springer International Publishing AG, part of Springer Nature 2018

A Caminha Muniz Neto, An Excursion through Elementary Mathematics, Volume III,

Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77977-5_1

1

The elementary theory of counting (configurations) has its foundations in the

following simple proposition, to which we will systematically refer as the bijective principle.

Proposition 1.1 If A and B are nonempty finite sets, then |A| = |B| if and only if

there exists a bijection f : A → B.

Proof First of all, assume that there exists a bijection f : A → B If |A| = n, we can take a bijection g : I n → A, so that f ◦ g : I n → B is also a bijection Hence,

|B| = n.

Conversely, suppose that |A| = |B| = n, with bijections g : I n → A and

h : I n → B Then h ◦ g−1: A → B is a bijection from A to B.  The following consequence of the bijective principle is sometimes referred to as

the additive principle of counting Before we state it, we recall that two sets A and

B are said to be disjoint if A ∩ B = ∅.

Proposition 1.2 If A and B are finite disjoint sets, then

|A ∪ B| = |A| + |B|.

A typical application of the former proposition consists in counting the number ofdifferent ways of choosing exactly one object out of two possible distinct kinds, suchthat there is a finite number of possibilities for each kind of object In the statement

of the proposition, the kinds correspond to the disjoint sets A and B, whereas the possibilities for each kind correspond to the elements of A and B.

An easy induction allows us to generalize the additive principle for n finite

pairwise disjoint sets This is the content of the coming

Corollary 1.3 If A1 , A2, , A n are finite pairwise disjoint sets, then

For what comes next, given two sets A and B, we let A \ B denote the set

A \ B = {x ∈ A; x /∈ B}

and say that A \B is the difference between A and B, in this order If B ⊂ A, and if

no danger of confusion arises, we shall sometimes refer to A\B as the complement

of B in A, in which case we denote it as B c , instead of A \ B.

The simple formula of the next corollary, which counts the number of elements

of the complement of a subset of a finite set, is an additional consequence of theadditive principle

1.1 The Bijective Principle 3

Corollary 1.4 If A is a finite set and B ⊂ A, then

|B| = |A| − |A \ B|.

Proof Since A = B ∪ (A \ B), a disjoint union, the additive principle gives

|A| = |B ∪ (A \ B)| = |B| + |A \ B|.

 The general philosophy behind the use of the previous corollary in problems ofcounting is this: suppose we wish to count the number of elements of a certain finite

set B, but do not know how to do it directly An interesting strategy is to search for

a finite set A ⊃ B such that we know how to count both |A| and |A \ B| Then, we apply the formula of the corollary to compute the desired number of elements of B.

Some concrete examples of this situation will be found in what is to come

Our next result generalizes Proposition1.1 and Corollary 1.4, computing thenumber of elements of a union of two finite sets Formula (1.1) below is known as

the principle of inclusion-exclusion for two finite sets, and will be generalized in

Sect.2.1(cf Theorem2.1)

Proposition 1.5 If A and B are finite sets, then

|A ∪ B| = |A| + |B| − |A ∩ B|. (1.1)

Proof Since A and B \ A are finite, disjoint and such that A ∪ B = A ∪ (B \ A), the

additive principle gives

|A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A|.

On the other hand, we also have the disjoint union

B = (B \ A) ∪ (A ∩ B),

so that, again from the additive principle,|B| = |B \ A| + |A ∩ B| Then, |B \ A| =

|B| − |A ∩ B|, and once we plug this formula into the above relation for |A ∪ B|

we get

|A ∪ B| = |A| + (|B| − |A ∩ B|).



For what comes next, recall that the cartesian product of sets A and B is the

set A × B whose elements are the ordered pairs (a, b) with a ∈ A and b ∈ B In

mathematical symbols,

A × B = {(a, b); a ∈ A, b ∈ B}.

It is also worth recalling that ordered pairs possess the following important property:

if (a, b), (c, d) ∈ A × B, then

(a, b) = (c, d) ⇔ a = c and b = d.

In this respect, see Problem5

The coming result, together with its subsequent corollary, are known as the

multiplicative principle or as the fundamental principle of counting.

Proposition 1.6 If A and B are nonempty finite sets, then A × B is also finite, with

Now, since f : A → A × {y j } given by f (x) = (x, y j )is a bijection (with

inverse g : A×{y j } → A given by g(x, y j ) = x), the bijective principle guarantees

that|A| = |A × {y j }| for 1 ≤ j ≤ m Hence, it follows from (1.2) that

In applications, we ought to invoke this version of the fundamental principle of

counting whenever we need to choose two objects simultaneously, such that one

of the objects is of one of two possible kinds and the other object is of the otherkind (and each kind comprises a finite number of possibilities) In the statement of

the proposition, the two possible kinds correspond to the sets A and B, whereas the possibilities for each kind correspond to the elements of A and B.

It is time we look at a concrete example

Example 1.7 How many natural numbers have two distinct nonzero algarisms, both

less that or equal to 5? How many of them are such that the first algarism is smallerthat the second one?

1.1 The Bijective Principle 5

Solution Let A = {1, 2, 3, 4, 5} The first part of the problem is clearly equivalent

to counting how many ordered pairs (a, b) ∈ A × A are such that a = b; the second part requires that a < b.

Let us start counting the number of ordered pairs (a, b) ∈ A × A, without

additional restrictions Since this is the same as counting the number of elements of

A × A, the multiplicative principle gives 5 × 5 = 25 possible pairs (a, b) In order

to count how many of them are such that a = b, let us use Corollary1.4: the number

of such pairs is obtained by discounting, from the total number of pairs, those for

which a = b Since there are 5 pairs (a, a) with a ∈ A, we conclude that there are

25− 5 = 20 pairs (a, b) with a = b.

For what is left, note that (a, b) is an ordered pair such that a < b if and only

if (b, a) is an ordered pair for which b > a; in other words, the correspondence

(a, b) → (b, a) is a bijection between the set whose elements are the ordered pairs

(a, b) ∈ A×A such that a < b and that formed by the ordered pairs (a, b) ∈ A×A for which a > b Hence, these sets have the same number of elements (namely,

ordered pairs); since they are disjoint and their union equals the set of pairs (a, b)∈

A × A such that a = b, it follows from Proposition1.2that the desired number of

For the subsequent discussion, we need to extend the concept of cartesian product

to an arbitrary finite number of finite nonempty sets To this end, given finite

nonempty sets A1, A2, , A n , let’s define their cartesian product A1× A2 ×

· · · × A nas the set of sequences (to which we shall sometimes refer to asn-tuples)

The next corollary brings an important elaboration of the multiplicative principle

Corollary 1.9 Let A1 , A2, , A k be finite nonempty sets with |A1| = n1, |A2| =

n2, , |A k | = n k Then, there are exactly n1n2 n k sequences (a1, a2, , a k ) with a j ∈ A j for 1 ≤ j ≤ k.

1In view of this definition, in principle we have two distinct definitions for the elements of A × B:

on the one hand, they consist of the ordered pairs (a, b) such that a ∈ A and b ∈ B; on the other, they are sequences (a, b) for which a ∈ A and b ∈ B Since the ordered pair (a, b) is defined

by (a, b) = {{a}, {a, b}} (cf Problem5) and the sequence (a, b) (with a ∈ A and b ∈ B) is the function f : {1, 2} → A ∪ B such that f (1) := a ∈ A and f (2) := b ∈ B, we come to the

conclusion that, although we have been using the same notation, they are distinct mathematical

objects However, for our purposes the identification of the ordered pair (a, b) to the sequence

(a, b)is totally harmless and will be done, from now on, without further comments.

Proof As we have told above, a sequence (a1, a2, , a k ) such that a1∈ A1, a2∈

A2, , a k ∈ A k is nothing but an element of the cartesian product A1× A2× · · · ×

A k Hence, the number of such sequences equals the number of elements of A1×

A2× · · · × A k, which in turn equals, by the previous corollary,|A1||A2| |A k| =

In words, the above corollary furnishes a method for counting in how many

(distinct) ways we can choose k objects in order but independently, with the first object being of type 1, the second being of type 2, , the k-th being of type k and

having at our disposal one or more possibilities for the choice of each type of object

Here again, the distinct types of objects correspond to the sets A1, , A k, whereasthe numbers of possibilities for each type of object correspond to the elements ofthe sets under consideration

As a special case of the situation of Corollary1.9, we can count in how many

ways it is possible to choose, in order but independently, k elements out of a set

of n elements, with repeated choices being allowed More precisely, we have the

following

Corollary 1.10 If |A| = n, then there are exactly n k sequences of k terms, all chosen from the elements of A.

In other words, we say that the above corollary counts how many arrangements

with repetition of k objects, chosen out of a set with n objects, there exist.

Yet another way of looking at the result of Corollary1.10is to start by recalling

that a sequence of k terms, all belonging to I n , is simply a function f : I k → I n;hence, what we established in that corollary was the fact that the number of distinct

functions f : I k → I n is exactly n k Later, we shall compute how many of suchfunctions are injective and how many are surjective

The coming example applies the circle of ideas above to a concrete situation For

its solution, the reader might find it helpful to recall the criterion of divisibility by 3,

which will be derived in Problem1, page162: the remainder of a natural number n

upon division by 3 equals that of the sum of its algarisms.

Example 1.11 Compute the quantity of natural numbers n, of ten algarisms and

satisfying the following conditions:

(a) n does not end in 0.

(b) n is divisible by 3.

Solution If n = (a1a2 a9a10) the decimal representation of n, then m =

(a1a2 a9) is a natural number of nine algarisms and a10 ∈ {1, 2, 3, , 9} Now, for a fixed m = (a1a2 a9), the criterion of divisibility by 3 assures

that each of the numbers (a1a2 a91), (a1a2 a94) and (a1a2 a97) leave

the same remainder upon division by 3, the same happening with the numbers

(a1a2 a92), (a1a2 a95) and (a1a2 a98), as well as with the numbers

(a1a2 a93), (a1a2 a96) and (a1a2 a99) Moreover, letting r1, r2 and r3

denote such common remainders, then r , r and r are pairwise distinct

1.1 The Bijective Principle 7

Hence, for each natural number m = (a1a2 a9)of nine algarisms, we have

exactly three natural numbers n = (a1a2 a9a10)which are divisible by 3 Since

a1 ∈ {1, 2, , 9} and a2, , a9 ∈ {0, 1, 2, , 9}, Corollary1.9guarantees thatthe total of numbers we want to compute is 9× 108× 3 = 27 × 108 

Problems: Sect 1.1

1 * Prove that the notion of number of elements of a nonempty finite set is a well

defined concept More precisely, prove that there exists a bijection f : I m → I n

if and only if m = n.

2 * Prove the additive principle of counting from the definition for the number of

elements of a finite set More precisely, prove that if A and B are nonempty,

finite disjoint sets, with|A| = m and |B| = n, then there exists a bijection

f : I m +n → A ∪ B.

3 * Prove Corollary1.3

4 * Let A and B be nonempty finite sets, with |A| = |B| Prove that a function

f : A → B is injective if and only if it is surjective.

5 * Given sets A and B and elements a ∈ A, b ∈ B, we formally define the

ordered pair (a, b) by letting2

(a, b) = {{a}, {a, b}}.

Use this definition to show that, for a, c ∈ A and b, d ∈ B, one has (a, b) =

Moreover, if B1, , B n are pairwise disjoint, prove that A × B1, , A × B n

are also pairwise disjoint

7 * Prove Corollary1.8

For the next problem, we define an ordered partition of a set A as a

sequence (A1, , A k ) of subsets of A satisfying the following conditions: (i)

A = A1∪ ∪ A k ; (ii) A1, , A k are pairwise disjoint In this case, we use

to say that A1, , A k (in this order) form an ordered partition of A into k of

its subsets.

8 Let n, k ∈ N and A be a finite set with n elements Show that there exist exactly

k n ordered partitions of A into k subsets A1, , A k

2 Such a definition is due to Kazimierz Kuratowski, Polish mathematician of the twentieth century.

9 In the cartesian plane, a pawn moves according to the following rule: being at

the point (a, b), he can go to one of the points (a + 1, b + 1), (a + 1, b − 1),

(a − 1, b + 1) or (a − 1, b − 1) If the pawn starts at the point (0, 0), how many distinct positions can he occupy after his first n moves?

10 (AIME—adapted) Given k, n∈ N, do the following items:

(a) For a given integer 1 ≤ j ≤ n, prove that there are exactly (n − j + 1) k

sequences formed by k elements of I n(possibly with repetitions) such that

the smallest term of the sequence is greater than or equal to j

(b) For a given integer 1≤ j ≤ n, prove that there are exactly (n − j + 1) k−

(n − j) k sequences formed by k elements of I n(possibly with repetitions),

such that the smallest term of the sequence equals j

(c) Prove that the sum of the smallest terms of all n k sequences of k elements

of I n(possibly with repetitions) equals 1k+ 2k + · · · + n k

11 (France) Let k ∈ N, A = {1, 2, 3, , 2 k } and X be a subset of A satisfying the following condition: if x ∈ X, then 2x /∈ X Find, with proof, the greatest possible number of elements of X.

1.2 More Bijections

In this section, with the elementary results of the previous section at our disposal, wepresent some instances of deeper applications of the bijective principle to establishthe equality of the numbers of elements of two finite sets

We start by computing, in two different ways, the number of subsets of a finite

set If A is any set, we let P(A) denote the power set of A, i.e., the family3

P(A) = {B; B ⊂ A}.

Given finite sets A and B, both with n elements, a bijection f : A → B naturally

induces a bijection ˜f : P(A) → P(B), defined for C ⊂ A by

3In Set Theory, a family is a set whose elements are also sets.

1.2 More Bijections 9

Hence, as a corollary to the bijective principle, if A and B are nonempty finite

disjoint sets, then

|A| = |B| ⇒ |P(A)| = |P(B)|. (1.3)

We shall also need another concept, which will be useful in future discussions

Definition 1.12 Let A be a nonempty set For B ⊂ A, the characteristic function

of B (with respect to A) is the function χ B : A → {0, 1}, defined by

χ B (x)= 0, if x / ∈ B

For example, if A = {a, b, c, d, e, f, g} and B = {c, f, g}, then χ B is the

function from A to {0, 1} such that

χ B (a) = χ B (b) = χ B (d) = χ B (e) = 0 and χ B (c) = χ B (f ) = χ B (g) = 1 For what follows, if A = {a1, , a n } is a set with n elements, whenever convenient we can associate to A the sequence (a1, , a n ); whenever we do that,

we shall say that (a1, , a n ) is an ordering of (the elements of) A or, sometimes, the natural ordering of A.

If B ⊂ A, a (combinatorially) more interesting way of looking at the acteristic function χ B of B with respect to A is to consider it as a sequence of

char-0’s and 1’s, with the positions of the 1’s (with respect to the natural ordering of

A ) corresponding to the elements of B For instance, let A = {a, b, c, d, e, f, g},

furnished with the ordering induced from the lexicographical (i.e., alphabetical) order; if B = {c, f, g}, then the characteristic function of B corresponds to the sequence (0, 0, 1, 0, 0, 1, 1).

More generally, let A = {a1, , a n}, furnished with the natural ordering For

B ⊂ A, the characteristic sequence of B in A is the sequence s B = (α1, , α n ),such that

α j = 0, if a j ∈ B /

We can finally compute the number of subsets of a set with n elements.

Theorem 1.13 A set with n elements has exactly 2 n subsets.

Proof Let A = {a1, , a n } be a set with n elements, furnished with its natural ordering, and let S be the set of sequences of n terms, formed by the elements of the

set{0, 1} By Corollary1.10, we have|S| = 2 n We will show that|P(A)| = 2 n,and to this end it suffices to show that the function

f : P(A) −→ S

B −→ s B

(i.e., the function which associates to each subset of A its characteristic sequence with respect to A) is a bijection For what is left to do, check that the function

f, whereas the existence of such a representation is equivalent to the surjectivity of

Proof Let F0 andF1 be the families of the subsets of I n with sum of elementsrespectively even and odd SinceF0andF1are disjoint and such thatF0∪ F1 =

P(A), it follows from the additive principle and Theorem1.13that

|F0| + |F1| = |P(A)| = 2 n

Thus, if we show that|F0| = |F1|, we will get |F0| = |F1| = 2n−1.

where B c = A \ B denotes the complement of B in A Since (B c ) c = A \ (A \ B) =

B , we have f (f (B)) = B for every B ⊂ A; hence, f ◦ f = Id P(A) , so that f is a

bijection (cf Example 6.40 of [8]) Now, the idea is to show that, if n = 4k + 1 or

n = 4k + 2, then f induces a bijection between F0andF1

Therefore, the restriction g of f to F0appliesF0intoF1, whereas the restriction

h of f to F1 appliesF1 intoF0 However, since g and h are clearly inverses of each other (for, f = f−1), it follows that g and h are also bijections, so that

The bijective principle is particularly useful to understand the properties of the

partitions of a natural number n Here and in all that follows, a partition of the

natural number n is a way of writing n as a sum of (one or more) not necessarily

distinct natural summands For example, the distinct partitions of 4 are

4= 1 + 3 = 1 + 1 + 2 = 1 + 1 + 1 + 1 = 2 + 2.

The coming example is due to L Euler

Example 1.15 (Euler) Prove that the number of partitions of a natural n in odd summands equals the number of partitions of n in distinct summands.

Proof Letting I n denote the set of partitions of n in odd summands and D nthe set

of partitions of n in distinct summands, it suffices to construct a bijection f : I n→

D n To understand how to define f , consider the following partition of 51 in odd

where a1, a3, a5, ≥ 0 and, from some natural number k on, the number a 2k+1of

summands equal to 2k+ 1 is 0 Then, we can write

n = a1· 1 + a3· 3 + a5· 5 + a7· 7 + · · · From here, in order to get a partition f (P ) of n into distinct summands, substitute each positive coefficient a 2k+1by its binary representation (cf Example 4.12 de [8])and, if 2lis one of the summands of such a representation, let 2l ( 2k + 1) be one of the summands in f (P ) (you can easily notice that this is exactly what we did in the

particular case of the partitions of 51)

By systematically proceeding this way, we claim that we get a partition of n into distinct summands Indeed, any two summands of f (P ) can be written as

2l1 ( 2k1+ 1) and 2 l2 ( 2k2+ 1);

if k1 = k2, then such summands are clearly different from one another; if k1 =

k2, then we must have l1 = l2, for 2l1 and 2l2 are two summands of the binary

representation of a 2k1+1= a 2k2+1 In any case, f is well defined.

To prove that f is a bijection, let’s define a function g : D n → I n(which will be

the inverse of f ) in the following way: let Q be the partition

n = (m11+ m12+ · · · + m 1t1) + (m31+ m32+ · · · + m 3t3)

+ (m51+ m52+ · · · + m 5t )+ · · ·

1.2 More Bijections 13

of n into distinct summands, where we have grouped, within each pair of parentheses, all summands with a single odd part More precisely, what we are saying is that m 2k +1,i= 2l i ( 2k + 1), for some nonnegative integer l i Then,

1 Compute the number of subsets of the set I n which contain n.

2 * (APMO) We are given a natural number n and a finite nonempty set A Show that there are exactly (2 n − 1) |A| sequences (A1, A2, , A n ), formed by subsets

of A such that A = A1∪ A2∪ ∪ A n

3 (Soviet Union) We are given n straight lines in the plane, in general position, i.e.,

such that no two of them are parallel and no three of them pass through the samepoint Compute the number of regions into which these lines divide the plane.For the next problem, we say that a familyF of subsets of I nis an intersecting

system if, for every distinct A, B ∈ F, we have A ∩ B = ∅.

4 (Bulgaria—adapted) LetF be an intersecting system in I n

(a) Give an example of such anF in which |F| = 2 n−1.

(b) Prove that|F| ≤ 2 n−1, for any suchF.

5 * Given n ∈ N, let I n be the family of subsets of I n with odd numbers of

elements If A = {x1, x2, , x 2k+1} ∈ I n , with x1 < x2 < · · · < x 2k+1,

we say that x k+1is the central element of A, and write x k+1 = c(A) In this

respect, do the following items:

(a) For 1≤ x1< x2< · · · < x 2k+1≤ n, show that the correspondence

{x1, x2, , x 2k+1} → {n + 1 − x 2k+1, , n + 1 − x2, n + 1 − x1}

establishes a bijection between the sets in I n having central elements

respectively equal to i and n + 1 − i.

(b) In Problem5, page20, we will show that|I n| = 2n−1 Use the result of item(a), together with this fact, to show that

rule(s) concerning the natural number n We do this in a two-step process: firstly,

we use the declaration of A n to get a recurrence relation for the sequence (a n ) n≥1,i.e., a relation of the form

a n = F (a1, , a n−1, n), (1.6)

where F is some function of n variables; secondly, we use the recurrence relation, together with algebraic and/or analytical arguments, to compute or estimate a n in

terms of n.

In this section we concentrate our efforts in the first step above, by examining

in detail some specific examples ranging in difficulty from almost trivial to realchallenging As for the second step, for the time being we assume that the reader isfamiliar with the solution of linear recurrence relations of order at most three andwith constant coefficients, as presented in Chapter 3 of [8] Chapters3and21willdevelop more powerful tools for the treatment of more general recurrence relations

We start with a recursive counting for the number os subsets of a finite set

Example 1.16 If A is a set with n elements, then A has exactly 2 nsubsets

Proof Let a n be the number of subsets of A (here, we are implicitly using (1.3)when we write|A| as a function of n) For a fixed x ∈ A, there are two kinds of subsets of A: those which contain x and those which do not.

1.3 Recursion 15

If B is a subset of A containing x, then B \ {x} is a subset of A \ {x}; conversely,

if B is a subset of A \ {x}, then B ∪ {x} is a subset of A containing x Since

such correspondences are clearly inverses of each other, we conclude that there are

as many subsets of A containing x as there are subsets of A \ {x}; however, since

|A \ {x}| = n − 1, it follows that there are exactly a n−1of such subsets of A On the other hand, since the subsets of A not containing x are precisely the subsets of

A \ {x}, we have exactly a n−1subsets of this second kind

The argument of the previous paragraph clearly gives a n = 2a n−1 for every

n > 1 Since we obviously have a1= 2, the formula for the general term of a GPgives

a n = a1· 2n−1= 2n

 Our next example provides a recursive approach to Problem3, page13

Example 1.17 (Soviet Union) We are given n straight lines in the plane, in general position, i.e., such that no two of them are parallel and no three of them pass through

the same point Compute the number of regions into which these lines divide theplane

Solution Let a kbe the number of regions into which the plane gets divided by some

k straight lines in general position We trace out one more straight line, say r, such that the k + 1 resulting lines are also in general position Since r intersects the k other lines in k distinct points, we conclude that r gets divided in k+ 1 intervals by

these k points In turn, each one of these k+ 1 intervals corresponds to exactly one

region, from the a k regions we had, that r splits into two new regions Therefore, when we trace out line r we extinguish k + 1 older regions and generate 2(k + 1) new regions Thus, if a k+1denotes the total number of regions we get after tracing

out line r, we conclude that

The coming example is perhaps the most celebrated of all elementary

applica-tions of recursive reasoning, being known as the Fibonacci problem.4

Example 1.18 A couple of rabbits starts generating descendants when is at least 2

months old When this is so, it generates a new couple of baby rabbits per month

If we start with a couple of just-born baby rabbits at month one, how many coupleswill we have after 12 months?

Solution Let F n denote the total number of couples of rabbits after n months, so that F1 = 1 and F2 = 1 (the first couple of descendants will be born only after 3

months, so will be counted just in F3)

Now, let’s look at the F k+2couples of rabbits we will have after k+ 2 months

There are two possibilities for such a couple: either it already existed after k+ 1

months, or else it was born at the (k + 2)-th month The first possibility amounts,

by definition, to a total of F k+1couples As for the second one, note that the couple

of rabbits under consideration descends from one of the couples that already existed

after k months; conversely, each couple that already existed after k months generates

a new couple of descendants that will be born at the (k + 2)-th month, so that this gives an additional number of F k couples of rabbits after k+ 2 months

Hence, the total numbers of couples of rabbits after k, k + 1 and k + 2 months

are related by the recursive relation

Example 1.19 Using the letters A, B and C, compute how many words of ten letters

(not necessarily with a meaning) one can form, so that there are no consecutiveconsonants

Solution For n ≥ 1, let a n be the number of words of n letters A, B, C satisfying the given condition on consonants For k ≥ 3, a word of k letters can finish in A,

B or C If it finishes in A, the k − 1 remaining letters form any of the a k−1word

of k − 1 letters and without consecutive consonants; if it finishes in B or C, then

4 After Leonardo di Pisa, also known as Fibonacci, Italian mathematician of the eleventh century Apart from its own contributions to Mathematics—as the problem we are presently describing— one of the greatest merits of Fibonacci was to help revive, in Middle Age Europe, the Mathematics

of Classical Antiquity; in particular, Fibonacci’s famous book Liber Abaci introduced, in Western

Civilization, the Hindu-Arabic algarisms and numbering system.

1.3 Recursion 17

the next to last letter must necessarily be an A (since we cannot have consecutive consonants), and the k − 2 initial letters form any of the a k−2words of k− 2 lettershaving no consecutive consonants

The reasoning of the previous paragraph gives us the recurrence relation

a k = a k−1+ 2a k−2, ∀ k ≥ 3.

Since a1= 3 and a2= 5 (the possible words of two letters and with no consecutive

consonants are AA, AB, AC, BA and CA), we successively compute a3 = 11,

a third order linear recurrence relation In this respect, perhaps the reader might find

it helpful to pause for a moment and review the content of Section 3.3 of [8].Alternatively, he/she can take a quick look at Problem5, page79, or at the material

of Chap.21(especially Theorem21.22.)

Example 1.20 For each set A = {x1, x2, , x m } of real numbers, with x1< x2<

Solution Let d n=∅=A⊂ I n (A) By examining some particular cases we easily

get d1= 1, d2= 4 and d3= 14 In general,

To analyse the last sum above, let A = A \ {n + 1} ⊂ I n , for each A ⊂ I n+1

such that n + 1 ∈ A Writing P nandI n for the families of subsets of I nwith evenand odd numbers of elements, respectively, we have

In Problem5(see also Problem7, page29) we shall prove that|P n | = |I n| =

2n−1 Assuming this fact for the time being, let’s separately consider the two sums

at the right hand side of (1.8):

(a) If A∈ P n , it easily follows from the definition of (·) that

Note that x k+1∈ I n is the central element of A∈ I n Hence, letting c(A)

denote the central element of A∈ I n, we obtain

Fig 1.1 The tower of Hanoi

game

3 A 2× n checkerboard must be tiled with dominoes (each of which is supposed

to have the shape of a 1×2 rectangle) If andenotes the number of such distincttilings, do the following items:

(a) Prove that a k+2= a k+1+ a k , for every integer k≥ 1

(b) Compute a n as a function of n.

4 Let n > 1 distinct circles be given in the plane, such that any two of them have

a common chord and no three of them pass through a single point Compute, as

a function of n, the number of regions into which the plane gets divided by the

circles

5 * We are given n ∈ N and a set A with n elements.

(a) Let a n and b n respectively denote the totals of subsets of A with even and odd numbers of elements Use a recursive argument to show that a n =

a n−1+ b n−1= b n

(b) Conclude that A has exactly 2 n−1subsets with an even number of elementsand 2n−1subsets with an odd number of elements.

The coming problem is a particular case of Kaplansky’s first lemma, which

will be dealt with, in general form, in Example1.28

6 Three nonconsecutive chairs are to be chosen out of a row of ten chairs Howmany are the ways of doing this?

7 The Tower of Hanoi game consists of three parallel equal rods, vertically

attached to the surface of a table, together with a pile of n disks of pairwise

unequal sizes, all having a hole in the center through which they can slide alongthe rods Figure1.1shows the initial configuration of the game for n= 5, withall of the disks piled at the leftmost rod, from the smallest disk at the top to thelargest at the bottom of the pile The purpose of the game is to move all disks

to the rightmost rod, possibly with the help of the central rod and subjected tothe following rule: at no moment a disk can be placed above a smaller one, in

any rod For general n, let a n denote the last number of movements needed tofinish the game Show that:

(a) a k+1= 2a k + 1, for every k ∈ N.

(b) a n= 2n − 1, for every n ∈ N.

8 * Given k, n ∈ N, the Stirling number of second kind5 S(n, k) is defined

as the number of ways of partitioning a set of n elements into k nonempty, pairwise disjoint subsets For example, S(3, 2)= 3, since

5 After James Stirling, Scottish mathematician of the eighteenth century With respect to Stirling numbers of second kind, see also Problem 4 , page 57

9 Compute, in terms of n ∈ N, the number of sequences of n terms, all of which

equal to 0, 1, 2 or 3 and having an even number of 0’s

10 A flag with n strips consists of n consecutive horizontal strips, each of which is

colored red, white or blue and such that any two adjacent strips have differentcolors

(a) Calculate how many are the distinct flags with n strips.

(b) If a n denotes the total number of flags with n strips and such that the first and last strips have different colors, show that a n+1+ a n = 3 · 2nfor every

n≥ 1

(c) Compute a n as a function of n, for every n≥ 1

11 (Bulgaria) A finite nonempty set A of positive integers is said to be selfish if

|A| ∈ A A selfish set A is minimal if A does not contain a selfish set different

from itself Do the following items:

(a) Prove that the number of minimal selfish subsets of I nthat do not contain

n coincides with the number of minimal selfish subsets of I n−1

(b) Given m ∈ Z and X ⊂ Z, let X+m denote the set X+m = {x+m; x ∈ X} Show that, for every integer n > 2, the correspondences

A → (A \ {n}) − 1 and B → (B + 1) ∪ {n}

are inverse bijections between the family of minimal selfish subsets of I n

that do not contain n and that of minimal selfish subsets of I n−2

(c) Show that I n has exactly F n minimal selfish subsets of I n , where F ndenotes

the n-th Fibonacci number.

For the next two problems, given a nonempty finite set A of real numbers, we let σ (A) and π(A) respectively denote the sum and the product of the elements

of A, with the convention that σ (A) = π(A) = a if A = {a}.

12 Given a nonempty finite set X ⊂ N, let a Xdenote the sum

∅=A⊂X

1

π(A) .

Do the following items:

(a) If m ∈ N \ X and Y = X ∪ {m}, show that a Y =m+1

(c) Show that a n = n for every n ∈ N.

13 (USA—adapted) For a nonempty finite set X ⊂ N, let b Xdenote the sum

∅=A⊂X

σ (A) π(A) .

(a) If m ∈ N \ X and Y = X ∪ {m}, show that

where a Xis the sum defined in the previous problem

(b) Given n ∈ N, write b n instead of b I n Show that

14 (France—adapted) We say that a set X of natural numbers is good provided

it satisfies the following property: for every natural number x, if x ∈ X then 2x / ∈ X For each natural k, let A k = {1, 2, 3, , 2 k } and b k be the greatest

number of elements which a good subset of A kcan have

(a) Show that b k = b k−2+ 2k−1for every k≥ 2

(b) For each natural n, compute b n as a function of n.

1.4 Arrangements, Combinations and Permutations 23

As an application of the ideas of the previous section, we use recursive arguments

to solve three specific, though very important, counting problems, namely, those of

counting the number of arrangements without repetitions, of permutations and of

combinations For what follows, we recall that I n = {1, 2, , n} for every n ∈ N.

We start by counting, in the coming result, the number of injective functionsbetween two nonempty finite sets

Proposition 1.21 Let n, k ∈ N If A is a set with n elements, then there are exactly

n(n − 1) (n − k + 1) injective functions f : I k → A.

Proof Let B be a set with m elements and let a kmdenote the number of injective

functions f : I k → B.

Evidently, a 1n = n (since there are exactly n possible choices for f (1) ∈ A) and,

in this case, the formula in the statement of the proposition is true

From now on, assume that k > 1 For a fixed x ∈ A, if f : I k → A is an injective function such that f (k) = x, then the restriction of f to I k−1is an injective function

˜

f : I k−1→ A \ {x} Conversely, given an injective function ˜ f : I k−1→ A \ {x},

we extend ˜f to an injective function f : I k → A by letting f (k) = x.

Since such operations of restriction and extension of injective functions areclearly inverses of each other, we conclude that there are as many injective functions

f : I k → A with f (k) = x as there are injective functions ˜ f : I k−1 → A \ {x} Thus, the number of such functions is exactly a k −1,n−1 However, since there are n possible choices for x ∈ A (for, |A| = n), we get the recurrence relation

which is valid for every natural n > 1.

Now, observe that a k1 = 0 for every k > 1 (for, if |A| = 1, there is no way of choosing distinct f (1) and f (2) in A) By induction, assume that a kn= 0 whenever

k > nand 1≤ n < m, where m > 1 is a natural number Given k, m ∈ N such that

k > m , we have k − 1 > m − 1, so that, by induction hypothesis, a k −1,m−1 = 0.But then, (1.9) gives a km = ma k −1,m−1= 0 Therefore, it follows by induction that

a kn = 0 for every k, n ∈ N such that k > n Yet in this case we have n−k+1 ≤ 0, so that n(n − 1) (n − k + 1) = 0 and the formula of the statement of the proposition

Since an injective function f : I k → A is simply a sequence of k pairwise distinct terms, all chosen from the elements of A, we can rephrase the above

proposition according to the following

Corollary 1.22 If |A| = n, then there are exactly n(n−1) (n−k +1) sequences

formed by k pairwise distinct elements of A.

We can apply this corollary to a given combinatorial situation whenever it asks us

to count how many are the ordered choices of k pairwise distinct elements of a given set A For this reason, from now on we shall say that the previous corollary counts

the number of arrangements without repetition of k pairwise distinct elements,

chosen from a given set of n elements.

Example 1.23 How many are the natural numbers of three pairwise distinct

algarisms?

Solution To choose a natural number with three pairwise distinct algarisms is the

same as to choose a sequence (a, b, c) such that a, b, c ∈ {0, 1, 2, , 9} are pairwise distinct and a= 0 Corollary1.4assures that, in order to count the number

of possible such sequences, it suffices to count the number of sequences (a, b, c), with pairwise distinct a, b, c ∈ {0, 1, 2, , 9}, and then to discount the number of such sequences of the form (0, b, c).

By the previous corollary, the number of sequences (a, b, c) with a, b, c ∈

{0, 1, 2, , 9} pairwise distinct but without the restriction a = 0 is 10 × 9 × 8 =

720 On the other hand, the number of such sequences of the form (0, b, c) is

9× 8 = 72 Hence, the desired result is 720 − 72 = 648  The counting of arrangements without repetition has a very important conse-quence, based on the following

Definition 1.24 A permutation of the elements of a nonempty set A is a bijection

f : A → A.

If A is finite and nonempty, it is not difficult to prove (see Problem4, page7)

that a function f : A → A is bijective if and only if it is injective Hence, letting

k = n in the formula of Proposition1.21, we immediately get our next result

Corollary 1.25 If |A| = n, then there are exactly n! permutations of the elements

of A.

Given a finite set A with n elements and a natural number k such that 0 ≤ k ≤ n, the coming result uses a recursive argument to compute the number of subsets of A with k elements each In order to properly state it, recall (cf Section 4.2 of [8]) that,

for nonnegative integers n and k with 0 ≤ k ≤ n, one defines the binomial number



k !(n − k)! .

1.4 Arrangements, Combinations and Permutations 25

Moreover, for 1≤ k ≤ n such numbers satisfy the recurrence relation



n k

Proposition 1.26 If A is a finite set with n elements and 0 ≤ k ≤ n, then A

= 1 Thus, let 1 ≤ k ≤ n and C n

k be the number of subsets of A with k

elements

For a fixed x ∈ A, there are two kinds of subsets of A with k elements: those which do not contain x and those which do contain x The former ones are precisely the k-element subsets of A \ {x}; since |A \ {x}| = n − 1, there are exactly C n−1

these subsets of k elements of A.

On the other hand, if B ⊂ A has k elements and x ∈ B, then B \ {x} ⊂ A \ {x} has k −1 elements; conversely, if B⊂ A\{x} has k−1 elements, then B∪{x} ⊂ A has k elements, one of which is x Since such correspondences are clearly inverses

of each other, we conclude that there are as many k-element subsets of A containing

x , as there are k − 1 element subsets of A \ {x}; thus, there are exactly C n−1

k

 Finally, since

(for A has exactly n subsets of 1 element each—

the sets{x}, with x ∈ A), an easy induction gives C n

k =n k

for 0≤ k ≤ n. 

In words, the previous proposition computes how many are the unordered choices

of k distinct elements of a set having n elements; one uses to say that such choices

are the combinations ofn objects, taking k at a time Also thanks to the former

proposition, one uses to refer to the binomial numbern

k



as “n chooses k”.

Example 1.27 When all diagonals of a certain convex octagon have been drawn,

one noticed that there were no three of them passing through a single interior point

of the octagon How many points in the interior of the octagon are intersection points

of two of its diagonals?

Solution Firstly, note that the condition on the diagonals of the octagon guarantees

that each one of the points of intersection we wish to count is determined by asingle pair of diagonals Hence, it suffices to count how many pairs of diagonals ofthe octagon intersect in its interior To this end, note that each 4-element subset of

the set of vertices of the octagon determines exactly two diagonals which intersect

in the interior of it; conversely, if two diagonals of the octagon do intersect in itsinterior, then the set of their endpoints is a 4-element subset of the set of vertices ofthe octagon Therefore, the bijective principle assures that there are as many pairs ofdiagonals intersecting in the interior of the octagon as there are 4-element subsets

of its set of vertices By Proposition1.26, then, the number of points of intersection

we wish to count equals8

Kaplansky’s first lemma.6

Example 1.28 (Kaplansky) Given n, k ∈ N, with n ≥ 2k − 1, show that the number

k -element subsets of I nwithout consecutive elements equalsn −k+1

k



Proof We first note that if A ⊂ I n has k elements, no two of which being consecutive, then n ≥ 2k − 1, for between any two elements of A there is at least one element of I n not belonging to A Hence, the assumption n ≥ 2k − 1 made

in the statement is natural and gives n − k + 1 ≥ k, so that the binomial number

the number of such sequences

To what is left to do, let’s first consider a sequence of 2(n − k) + 1 terms, in which the n − k terms with even indices are equal to 0:

k

possible choices for such positions Finally, once weerase the not chosen positions, we are left with the characteristic sequence of a setsatisfying the conditions of the problem Conversely, it is immediate that every suchcharacteristic sequence can be obtained as above, so that the answer to our problem

isn −k+1

k



Now, we are given natural numbers k and n and nonnegative integers n1, , n k

satisfying n = n1+· · ·+n k We also have k distinct types of objects, say a1, , a k,

6There is also a Kaplansky’s second lemma, which will be the object of Problem12 Kaplansky devised these two results in order to solve Lucas’ problem—cf Problem 10 , page 41

1.4 Arrangements, Combinations and Permutations 27

and would like to count how many sequences of n terms have exactly n1terms equal

to a1, , n k terms equal to a k Yet in another way, we say that such a sequence is

a permutation with repeated elements of n objects, each of which being of one of

the types a1, , a k and with n1objects of type a1, , n k objects of type a k Let’snow look at an equivalent problem

If f : I n → {a1, , a k} is a sequence satisfying the conditions of the previous

paragraph (i.e., with n1terms equal to a1, , n k terms equal to a k), then, takinginverse images, we obtain

I n = f−1(a1) ∪ ∪ f−1(a k ).

At the right hand side above, the sets f−1(a j ), 1≤ j ≤ k, are pairwise disjoint and

such that|f−1(a j ) | = n jfor 1≤ j ≤ k; we say that the union at the right hand side

is the partition of I n induced by f Conversely, associated to a partition of I n of

the form I n = A1∪ ∪ A k, with|A j | = n j for 1≤ j ≤ k, we have the sequence

f : I n → {a1, , a k } such that f (A j ) = {a j } for 1 ≤ j ≤ k Therefore, the

bijective principle guarantees that a problem equivalent to the one of the previous

paragraph is that of counting the number of distinct partitions of a set of n elements into k pairwise disjoint subsets A1, , A k, under the restriction that|A1| = n1, ,

denotes the number of partitions

of A into sets A1, , A k , with |A j | = n j for 1 ≤ j ≤ k, then

n = n1+ · · · + n l of n in positive integers.

The bijective principle assures that, for each subset A l of A, with |A l | = n l,

there are as many partitions of A as in the statement of the proposition as there are

partitions

A \ A l = A1∪ ∪ A l−1,

of A \ A l into l − 1 subsets A1, , A l−1, with|A j | = n j for 1≤ j ≤ l − 1.

Since |A \ A l | = n − n l , the total number of such partitions of A \ A l is,

choosing the subset A l of A, the fundamental principle of counting gives the



,

where the right hand side denotes the usual binomial number Such and equality

reflects the fact that choosing a k-element subset out of a set with n elements is the same as partitioning such a set into two subsets, one with k elements and the other with n − k elements.

Problems: Sect 1.4

1 If n ∈ N and A is a set with n elements, prove that there are exactly n k − n(n − 1) (n − k + 1) sequences of k terms, all taken out of A but not pairwise

distinct

2 * Do the following items:

(a) Use a combinatorial argument to prove the formula for binomial expansion:

1.4 Arrangements, Combinations and Permutations 29

3 Given n, k∈ N, use the result of Proposition1.29, together with item (b) of the

previous problem, to compute the number of ordered partitions of a set A, with

|A| = n, into k subsets A1, , A k(cf Problem8, page7)

4 Prove that the Stirling number of second kind S(n, k) (cf Problem8, page20)can be computed with the aid of the following formula:

5 (IMO—shortlist) Given a permutation (a1, a2, , a n ) of I n , we say that a j is

a local maximum if a j is greater than its neighbors (for j = 1, this means that

a1> a2; for j = n, that a n > a n−1) Compute the number of permutations of

I nhaving exactly one local maximum

6 (OCM) We are given n ≥ 6 points on a circle , and draw alln

2

chords joiningtwo of them We further assume that no three of such chords pass through a

single point of the corresponding open disk D bounded by  Compute the

number of distinct triangles satisfying the two following properties:

i their vertices belong to D.

ii their sides lie on three of then

2

drawn chords

7 * Use combinations to show that a set with n elements has exactly 2 n−1subsetswith an even number of elements and 2n−1 subsets with an odd number ofelements

8 Compute, in terms of n ∈ N, the number of sequences of n terms, all of which

equal to 0, 1, 2 or 3 and having an even number of 0’s

9 * Given n, k ∈ N, show that the equation x1+ · · · + x k = n has exactlyn +k−1

k−1

nonnegative integer solutions

10 * Given n, k ∈ N, show that the equation x1+ · · · + x k = n has exactlyn−1

k−1

positive integer solutions

11 The purpose of this problem is to give another proof of Kaplansky’s first lemma(cf Example1.28) To this end, do the following items:

(a) If A = {a1, , a k } ⊂ I n is a set without consecutive elements, let x1 =

a1− 1, x k+1= n − a k and x j = a j − a j−1− 2 for 2 ≤ j ≤ k Show that

x1, , x k+1solves, in nonnegative integers, equation x1+ · · · + x k+1 =

n − 2k + 1.

(b) If x1, , x k+1is a solution of equation x1+ · · · + x k+1= n − 2k + 1 in nonnegative integers, show how to use it to get a set A = {a1, , a k } ⊂ I n

without consecutive elements

(c) Obtain Kaplansky’s first lemma from the results of items (a) and (b) andfrom Problem9

12 * Prove Kaplansky’s second lemma: given n, k ∈ N, with n ≥ 2k, the number

of k-element subsets of I nwithout consecutive elements, and considering 1 and

nto be consecutive, is equal to