quantities of odd and even elements is equal to2n
n
.
5. A battalion commander asked volunteers to compose 11 patrols, all of which formed by the same number of men. If each man entered exactly two patrols and each two patrols had exactly one man in common, compute the numbers of volunteers and of members of each patrol.3
6. (IMO) Given n, k ∈ N, let pn(k)denote the number of permutations of In having exactlykfixed points. Prove thatn
k=0kpn(k)=n!.
7. (TT) The set of natural numbers is partitioned intominfinite and nonconstant arithmetic progressions, of common ratiosd1,d2, . . . ,dm. Prove that
1 d1 + 1
d2 + ã ã ã + 1 dm =1.
The result of the coming problem is known asSperner’s lemma, after the German mathematician of the twentieth century Emanuel Sperner.
8. A triangleABCis partitioned into a finite number of smaller triangles, in such a way that any two of them are either disjoint, have exactly a common vertex or exactly a common side. Such a partition generates a certain number of points lying in the interior or on the sides ofABCand which are vertices of smaller triangles. We call themvertices of the partition. Randomly label the points lying on sideAB(and different fromAorB) as “A” or “B”; accordingly, randomly label the points lying on sideAC (and different fromAorC) as “A” or “C”, and those lying on sideBC(also different fromB orC) as “B” or “C”. Then, label each of the remaining vertices of the partition (all lying in the interior ofABC) as A,B or C, also randomly. Prove that at least one of the smaller triangles will be labelled asABC.
9. (Russia) A senate has 30 senators, any two of which are either friends or enemies. It is also known that each senator has exactly six enemies. If each three senators form a commission, compute the number of those commissions such that its members are either pairwise friends or pairwise enemies.
10. We are given a convexn-gon in the plane. Prove that it is possible to choose n−2 points in the plane such that every triangle formed by three vertices of the givenn-gon has exactly one of the chosen points in its interior.
2.3 Equivalence Relations and Counting
LetAbe a fixed nonempty set, and recall that arelationinAis a subset of the cartesian productA×A.
3For another approach to this problem, see Problem21, page134.
We say that a relationRinAisreflexiveifaRafor everya∈A;symmetricif, whenevera, b∈A, we have thataRb⇒bRa;transitiveif, whenevera, b, c∈A, we have thataRb andbRcimplyaRc. Finally, anequivalence relationinAis a relationRinAwhich is simultaneously reflexive, symmetric and transitive. In Mathematics, equivalence relations are customarily denoted by∼, instead ofR(i.e., by writinga∼binstead ofaRb); from now on, we shall stick to this usage without further comments.
Examples 2.14
(a) LetFbe a partition of a nonempty setA. The relation∼inA, defined by a ∼b⇔ ∃B∈F;a, b∈B,
is of equivalence. We say that such an equivalence relation is the oneinduced by the partitionF.
(b) LetAandBbe nonempty sets andf :A→Bbe a given function. Relation∼ inA, defined by
a ∼b⇔f (a)=f (b)
is of equivalence. We say that such a relation isinducedbyf. The following definition is central to all that follows.
Definition 2.15 LetAbe a nonempty set and ∼be an equivalence relation inA.
Fora∈A, theequivalence classofais the subsetaofAdefined by
a= {x∈A;x∼a}. (2.11)
If ∼ is an equivalence relation in a nonempty set A and a ∈ A, then the equivalence class of a is clearly nonempty, for a ∈ a. Our first result analyses the equivalence classes of a generic equivalence relation.
Proposition 2.16 Let∼be an equivalence relation in a nonempty setA. Fora, b∈ A, one has
a ∼b⇔a∩b= ∅ ⇔a=b.
In particular, the equivalence classes of∼form a partition ofA.
Proof We shall show thata∼b⇒a∩b= ∅ ⇒a =b⇒a∼b:
• Ifa∼b, thena∈band, hence,a∈a∩b; therefore,a∩b= ∅.
• Ifc∈a∩b(i.e., ifa∩b= ∅), thenc∼aandc∼b; hence, the symmetry of∼ guarantees thata∼candc∼band, thus, its transitivity impliesa∼b. We shall now show thata ⊂b(the converse inclusion being totally analogous): ifx ∈a, thenx ∼a; however, sincea ∼b, it follows once more from the transitivity of
∼thatx∼b, i.e.,x ∈b.
2.3 Equivalence Relations and Counting 51
• Ifa=b, thena∈aimpliesa ∈b. But this is the same as saying thata ∼b.
If∼is an equivalence relation in a nonempty setA, it follows immediately from the previous proposition that the familyA/∼, defined by
A/∼ = {a; a∈A} (2.12) (i.e., the family of the equivalence classes of the elements ofAwith respect to∼), is a partition ofA. Such a family is called thequotient setofAby∼, and the function
π :A−→A/∼
a −→ a (2.13)
is theprojection functionofAonA/∼.
Given an equivalence relation∼in a nonempty setA, a subsetBofAis said to be asystem of distinct representatives(we abbreviateSDR) for∼if the following conditions are satisfied:
(a) For alla, b∈B, eithera=bora∼b.
(b) For everya ∈A, there existsb∈Bsuch thata ∼b.
In words, a SDR for an equivalence relation∼inAis a subset ofA formed by pairwise nonrelated elements such that every element ofAis related to exactly one of them. Yet in another way, ifBis a SDR for the equivalence relation∼inA, then
A/∼= {x;x ∈B}. (2.14)
The following proposition provides an effective way of constructing SDR for equivalence relations.
Proposition 2.17 Let ∼be an equivalence relation in a nonempty set A. If f : A/∼ →Ais a function such thatf (a)∈afor everya ∈A, then Im(f )is a SDR for∼.
Proof LetB = Imf. Ifa, b ∈ B, saya = f (α)andb =f (β), withα, β ∈ A, then the hypothesis onf assures thata∈αandb∈β. Ifa∼b, then, sinceα∼a andb∼β, we have by transitivity thatα∼β; thus,α=βand, hence,
a=f (α)=f (β)=b.
In order to verify the validity of the second condition in the definition of a SDR, takea ∈Aand letb=f (a). Thenb∈ Im(f )and the definition off givesb∈a, i.e.,a∼b.
In the notations of the previous proposition, we say thatf is achoice function for the equivalence relation∼. In words, the role off is to choose one element out of each equivalence class inA/∼.
Concerning the role of equivalence relations in counting problems, we have finally arrived at the result we are interested in.
Proposition 2.18 Let∼be an equivalence relation in a finite nonempty setA. If all of the equivalence classes ofAwith respect to∼have the same numberk of elements, then there are exactly|A|/kdistinct equivalence classes.
Proof If{a1, . . . , am}is a SDR with respect to∼, thenA = m
j=1aj, a disjoint union. Hence,
|A| = m j=1
|aj| = m j=1
k=km, so thatm= |A|/k.
As a first example of application of the previous proposition, we shall now reobtain the formula for the number of k-element subsets of a given set A, ofn elements (the following reasoning is a slight variation of that in Example2.8, the major change being in the adopted viewpoint). To this end, let 1≤k≤nandSk(A) be the set of sequences ofkdistinct terms, all belonging toA. We already know that
|Sk(A)| =n(n−1) . . . (n−k+1)= n! (n−k)!. Let∼be defined inSk(A)by
(a1, . . . , ak)∼(b1, . . . , bk)⇔ {a1, . . . , ak} = {b1, . . . , bk}. (2.15) It’s immediate to check that ∼is of equivalence, and such that the equivalence class of a sequence(a1, . . . , ak)is the set of all of its permutations. Hence, such an equivalence class has preciselyk!elements, and Proposition2.18assures that the number of distinct equivalence relations is equal to
|Sk(A)|
k! = n!
k!(n−k)! = n
k
.
Finally, note that (2.15) allows us to identify the equivalence classes with the k- element subsets ofA, so that there are exactlyn
k
of such subsets.
Another interesting situation is the following: in the setS(A)of the permutations of a finite nonempty setA, we define an equivalence relation∼in the following way:
(a1, . . . , an)∼(b1, . . . , bn)
2.3 Equivalence Relations and Counting 53
(2.16)
∃1≤k≤n;bj =
aj+k, if 1≤j ≤n−k aj+k−n,ifn−k < j ≤n .
In other words,(a1, . . . , an)∼(b1, . . . , bn)means that, for some 1≤k ≤ n, one has
(b1, . . . , bn)=(ak+1, ak+2, . . . , an−1, an, a1, . . . , ak).
It’s immediate to verify that ∼is actually of equivalence inS(A). Moreover, for each sequence(a1, . . . , an)∈ S(A), its equivalence class(a1, . . . , an)contains exactlynsequences; more precisely,
(a1, . . . , an) = {(a1, a2. . . , an−1, an), (a2, a3, . . . , an, a1) (a3, . . . , an, a1, a2), . . . , (an, a1, a2, . . . , an−1)}.
With respect to the equivalence relation just described, the equivalence class of a sequence (a1, . . . , an)is said to be the circular permutationof the sequence (a1, . . . , an). In particular, it follows from the general discussion on equivalence relations that all of the sequences
(a1, a2. . . , an−1, an), (a2, a3, . . . , an, a1), . . . , (an, a1, a2, . . . , an−1) define the same circular permutation.
With the concept of circular permutation at our disposal, we now have the following result.
Proposition 2.19 The number of distinct circular permutations of a set of n elements is(n−1)!.
Proof IfAis a set havingnelements, we wish to count the number of equivalence classes of S(A) with respect to the equivalence relation just described. Since
|S(A)| = n! and each equivalence class has exactlyn elements, it follows from Proposition2.18that there are precisely nn! =(n−1)!distinct equivalence classes (i.e., circular permutations).
For what comes next, we say that two sets AandB are incomparable with respect to inclusion4 if A ⊂ B and B ⊂ A. The result below brings a deeper application of the ideas of this section, and is due to the Hungarian mathematician Béla Bollobás. Our discussion follows [25].
4This terminology comes from the fact that the inclusion relation is apartial orderin the family of all subsets of a given set. In this respect, see the discussion at Sect.4.3.
Theorem 2.20 (Bollobás) IfFis a family of subsets ofIn, pairwise incomparable with respect to inclusion, then
A∈F
n
|A| −1
≤1.
Proof LetP be the set of permutations(a1, . . . , an)ofInsuch that, for some 1≤ k≤n, we have{a1, . . . , ak} ∈F. Being a set of permutations ofIn, it is immediate that|P| ≤n!.
We consider inPthe relation∼, given by
(a1, . . . , an)∼(b1, . . . , bn)⇔ {a1, . . . , ak} = {b1, . . . , bk} ∈F,
for some integer 1≤ k≤ n. We claim that∼is of equivalence. Indeed, it is clear that∼is reflexive and symmetric. In order to show that it is transitive, take a third permutation(c1, . . . , cn)inP and assume that{a1, . . . , ak} = {b1, . . . , bk} ∈ F and{b1, . . . , bl} = {c1, . . . , cl} ∈F, for some integers 1≤ k, l≤ n. Without loss of generality, we can also assume thatk≤l. However, ifk < l, then
{a1, . . . , ak} = {b1, . . . , bk} ⊂ {b1, . . . , bl} = {c1, . . . , cl} ∈F,
a strict inclusion; this would contradict the fact that the sets in F are pairwise incomparable with respect to inclusion. Therefore,k=land, thus,
{a1, . . . , ak} = {b1, . . . , bk} = {c1, . . . , ck} ∈F, i.e.,(a1, . . . , an)∼(c1, . . . , cn).
We now fix a permutation(a1, . . . , an)∈P, with{a1, . . . , ak} =A∈F. Then, (b1, . . . , bn) ∼ (a1, . . . , an)if and only if{b1, . . . , bk} = A, i.e., if and only if (b1, . . . , bk)is a permutation of the elements ofA(and, hence,(bk+1, . . . , bn)is a permutation of the elements ofIn\A); since we can permute the elements ofAand ofIn\Ain exactly|A|!and(n−|A|)!ways, respectively, we conclude that there are exactly|A|!(n− |A|)!permutations(b1, . . . , bn)which are equivalent to the fixed permutation(a1, . . . , an)∈P. Yet in another way, ifAis the set inFassociated to (a1, . . . , an)∈P, then the equivalence class of(a1, . . . , an)inP contains exactly
|A|!(n− |A|)!permutations.
Finally, by Proposition 2.16, the quotient set P/ ∼of equivalence classes of P with respect to∼forms a partition of P. However, since distinct equivalence classes are associated to distinct setsA∈ F and every setA∈ F gives rise to an equivalence class, it follows that
n! ≥ |P| =
(a1,...,an)∈P/∼
#(a1, . . . , an)=
A∈F
|A|!(n− |A|!).
2.3 Equivalence Relations and Counting 55 This is the same as
A∈F
|A|!(n− |A|!) n! ≤1, which in turn is equivalent to the desired inequality.
For our next example we first need to recall some simple facts concerning the notion of divisibility of integers. Further details can be found in Chap.6.
Givena, b ∈ Z, withb =0, we say thatbdividesa, or thatais a multiple of b, if ab is an integer. Letting ab = c, this is the same as saying thata = bc, with c∈ Z. For instance, 7 divides 28, for 287 =4 ∈ Z. Ifbdividesa, we writeb | a;
otherwise, we writeb aand say thatbdoes not dividea. Ifa is not a multiple of b, thedivision algorithm5shows that there exist unique integersq andr such that a=bq+rand 0≤r <|b|.
We now introduce a whole class of equivalence relations inZ, which will reveal itself to be of paramount importance to the material of Chaps.10–20.
For a given integern >1, we say that two integersaandbarecongruent modulo nifndividesa−b. Ifa andbare congruent modulon, we writea ≡ b (modn);
otherwise, we writea ≡ b (modn). For instance, we have 1 ≡ −5(mod 3), since 3| (1−(−5)), but−3≡7(mod 4), since 4(−3−7). Note that ifa =nq+r, with 0≤r < n, thenn|(a−r), so thata≡r (modn). Therefore, every integer is congruent, modulon, to precisely one of 0, 1, . . . ,n−1.
Example 2.21 Letn >1 be a given integer. The relation∼, defined inZby a∼b⇔a≡b (modn),
is an equivalence relation in Z, which is known as the relation of congruence modulon.
Proof Reflexivity of∼follows from
a ∼a ⇔a ≡a (modn)⇔n|(a−a), which is clearly true. On the other hand, since
a∼b⇔a≡b (modn)⇔n|(a−b), andn|(a−b)⇔n|(b−a), we have that
a∼b ⇒n|(b−a)⇒n|(b−a)
⇒b≡a (modn)⇒b∼a.
5For a formal definition of analgorithm, see the footnote at page116.
Finally, for the transitivity of∼, ifa ∼ b andb ∼ c, thenn | (a −b) and n|(b−c), i.e., a−nbandb−nc are both integers; hence, the number
a−c
n =a−b
n +b−c n
is also an integer, which in turn guarantees thatn| (a−c). But this is the same as saying thata≡c (modn), so thata ∼c.
For the coming example, given an integerm > 1, a setA ⊂ Im and another integer 0≤r≤m−1, we letA+rdenote the set
A+r= {a+r (modm);a ∈A}. Thus,
A+r = {x+r;x ∈Aand 1≤x≤m−r}∪
∪ {x+r−m;x ∈Aandm−r < x≤m}. For instance, ifm=7,A= {1,3,5}andr=4, then
A+r= {1+4,3+4,5+4(mod 7)} = {5,7,2}.
Example 2.22 (Austrian-Polish) Letk andnbe given natural numbers, such that 3kand 1≤k≤3n. Prove that the setI3nhas exactly133n
k
subsets ofkelements, such that the sum of its elements is a multiple of 3.
Solution LetF be the family ofk-element subsets ofI3n. Consider a relation∼ inF be defining A ∼ B if and only if there exists an integer 0 ≤ j ≤ 2 such thatB = A+j modulo 3. It is immediate to check (cf. Problem2) that∼is an equivalence relation, and that the equivalence class of a subsetA of In contains exactly the three setsA,A+1 andA+2 (these last two modulo 3).
On the other hand, lettingS=
x∈Ax, we obviously have
x∈A+1
x ≡S+k (mod 3) and
x∈A+2
x ≡S+2k (mod 3).
However, since 3k, we conclude that the three numbersS,S+kandS+2kare pairwise incongruent modulo 3. But since every integer is congruent to precisely one of the integers 0, 1 or 2, modulo 3, we conclude that exactly one ofS,S+k andS+2kis divisible by 3. This is the same as saying that exactly one of the sets A,A+1 andA+2 has sum of elements equal to a multiple of 3. Hence, there are as manyk-element subsets ofI3nwhose sum of elements is divisible by 3 as there are equivalence classes ofF with respect to ∼. But since each such equivalence class contains three elements, Proposition2.18assures that the number of distinct equivalence classes is precisely 133n
k
.
2.3 Equivalence Relations and Counting 57
Problems: Sect. 2.3
1. In each of the items of Example2.14, prove that the given relations are, indeed, of equivalence. Also, identify the corresponding equivalence classes.
2. * Givenk, m∈ N, define a relation∼in the family ofk-element subsets ofIm by lettingA ∼B if and only if there exists 0 ≤ j ≤ 2 for whichB = A+j modulo 3. Prove that∼is an equivalence relation.
3. Show that the number of distinct ways of partitioning a set ofabelements intoa sets ofbelements each is equal to a(ab)!(b!)!a.
4. Givenn, k ∈ N, with 1≤ k ≤ n, letS(n, k)denote the corresponding Stirling number of second kind (cf. Problem8, page20) and letAbe a finite set withn elements. Do the following items:
(a) Show that every equivalence relation on A, with exactly k equivalence classes, can be obtained from a surjective functionf :A→B, with|B| =k, as prescribed by item (b) of Example2.14.
(b) Show that the number of equivalence relations on A having exactly k equivalence classes is equal toS(n, k).
(c) Conclude, from the previous items, that S(n, k)= 1
k! k j=0
(−1)j k
j
(k−j )n.
For the next problem, recall that a natural numberp >1 is said to beprimeif 1 andpare the only positive divisors ofp. We also recall (see Sects.6.2and6.3 for details) that ifa, b∈Nare such thatp|(ab)butpa, thenp|b.
5. We are given a prime numberpandmpairwise distinct colors. Prove that there are exactlympp−m+mdistinct necklaces, each of which formed bypbeads, each bead being of one of themgiven colors (there may well be beads of repeated colors). Then, use this counting to establishFermat’s little theorem6: ifp is prime and m is a natural number which is not divisible byp, then mp−1 ≡ 1(modp).
6. (IMO shortlist) Givenn∈N, we say that a sequence(x1, x2, . . . , xn), withxj ∈ {0,1}for 1 ≤ j ≤ n, isaperiodicif there does not exist a positive divisor d of nsuch that the sequence is formed by the juxtaposition of nd copies of the subsequence(x1, . . . , xd). Ifan denotes the number of aperiodic sequences of sizen, prove thatn|an.
7. Prove the following theorem of Sperner: ifFis a family of subsets ofIn, pairwise uncomparable with respect to inclusion, then
6After Pierre S. de Fermat, French mathematician of the seventeenth century. For a different proof of this result, see Sect.10.2.
|F| ≤ n
n/2
,
with equality ifF= {A⊂In; |A| = n/2}.
8. This problem generalizes Problem19, page31. To this end, letn∈N,pbe prime andf :In→Inbe such thatf(p)=Id, withf(p)standing for the composition off with itself,ptimes.
(a) Show thatf is a bijection and, if∼is the relation defined inInby a∼b⇔b=f(j )(a),∃j ∈Z,
then∼is an equivalence relation.
(b) Each equivalence class ofInwith respect to∼has either 1 orpelements. In particular, iff has exactlykfixed points, thenp|(n−k).
(c) Conversely, given 0≤k≤nsuch thatn−k=pl, withl∈Z+, show that for each choice of elementsx1, . . . , xk∈Inand for each partition
In\ {x1, . . . , xk} =A1∪. . .∪Al
ofIn\ {x1, . . . , xk}into pairwise disjointp-element sets, there are exactly (p−1)!l
functionsf as in the statement, having preciselyx1, . . . , xk as their fixed points and satisfyingf (Ai)=Ai for 1≤i≤l.
(d) Conclude that the number of functionsf :In →Insuch thatf(p) =Id is given by
0≤k≤n p|(n−k)
n k
ã
n−k p, . . . , p
(n−k)/p
(p−1)!n−kp
=
0≤k≤n p|(n−k)
n! k! ãpn−pk
.
For the coming problem, recall that a family F of subsets of In is an intersecting systemofInif, for allA, B∈F, we haveA∩B = ∅.
9. The purpose of this problem is to prove the famousErdửs-Ko-Rado theorem7: for positive integerskandn, with 1 ≤ k ≤ n, ifF is an intersecting system formed byk-element subsets ofIn, then
|F| ≤
n k
, if k > n/2 n−1
k−1
, if k≤n/2 .
To this end, do the following items:
7After the Hungarian, Chinese and German twentieth century mathematicians Paul Erdửs, Chao Ko and Richard Rado, respectively.