Chapter 2
Combinatorics
La Combinatoire est l’´etude:
• d’une configuration
• d’une configuration inconnue
• de d´enombrement exact de configurations
• de d´enombrement approch´e de configurations
• d’´enum´eration de configurations
• d’optimization
Claude Berge Principes de Combinatoire In this chapter on combinatorics, we start from the elementary rules of counting, then study permutations and combinations, bi- nomial coefficients, binomial theorem, multinomial coefficients, multinomial theorem, Stirling numbers of the first and the sec- ond kind, Bell numbers, the Principle of Inclusion and Exclusion (simple and weighted versions), some applications of the Principle of Inclusion and Exclusion to number theory and the theory of permanents, generating function techniques and recurrence rela- tions, Bernoulli numbers, Catalan numbers, and an algorithm for generating all the subsets of a given finite set.
2.1 What Is Combinatorics?
Combinatorics can be described as the study of properties offinite sets andfinitestructures (see [2]). Unless otherwise stated, all sets in this chapter are finite.
Combinatorics is traditionally associated with the permutation and combination of a finite number of objects. Combinatorics deals with the following two types of problems: Existence problems and enumeration problems (see [1]).
In the existence problem, the following question is raised:
Does there exist a particular special structure or situation or phenomenon? If the answer to the existence problem is affirma- tive, then we raise the second question: How many such special structures exist? This is the enumeration problem. In fact, the second problem of enumeration can be viewed as ageneralization of the existence problem. This is because, the existence of a spe- cial structure has an affirmative answer if and only if the number of such special structures is at least one.
The following examples illustrate the existence and enumera- tion problem in combinatorics.
Example 2.1.1 (Enumeration problem):
Consider a set A consisting of 4 integers 1,2,3,4. Symbolically, A = {1,2,3,4}. Find the number of subsets of A consisting of just two elements. This is an enumeration problem. Here the existence problem is trivially solved by “exhibiting” a subset of A with two elements (for example, {1,2} is a 2-element subset of A.)
There are six subsets of A containing exactly 2 elements, namely,{1,2}, {1,3, }, {1,4}, {2,3}, {2,4}, {3,4}. Example 2.1.2 (Existence problem: Checkerboard problem): Consider the familiar 8×8 chess board. From this chess board, we remove two diagonally opposite squares. The board thus obtained by the removal of two opposite squares is called acheckerboard. It has exactly 62 squares, the removed squares are marked with ×,
“B” stands for a black square and “W” stands for a white square
2.1. WHAT IS COMBINATORICS? 49 (see Table 2.1).
Table 2.1: Checkerboard
× B W B W B W B
B W B W B W B W
W B W B W B W B
B W B W B W B W
W B W B W B W B
B W B W B W B W
W B W B W B W B
B W B W B W B ×
Table 2.2: Dominoes
W B
Example 2.1.3:
The question is the following: Is it possible to cover the entire checkerboard by using thirty-one 1× 2 dominoes? This is the existence problem. The answer is “no” as the following simple
“parity” argument proves:
In the conventional chess board, the squares are colored alter- natively black and white. Hence there are 32 black squares and 32 white squares. Note that the two diagonally opposite squares are of the same color (either both black or both white). Suppose we have removed two white squares. Hence, our checkerboard has 30 white squares and 32 black squares. But a 1×2 rectangle of domino covers exactly one black square and one white square. The given 31 dominoes can cover 31 black squares and 31 white squares.
But the checkerboard has an unequal number of black and white squares. Hence a complete covering of the checkerboard with 31 1×2 dominoes is impossible.
On the other hand, if we are given 32 dominoes of 1×2 rectan- gle, we can easily cover the entire chess board with 32 dominoes.
In this case, we are led to the enumeration problem: How many such coverings are possible?
Example 2.1.4 (Existence problem: Euler’s 36 officers problem): Euler’s 36 officers problem and the fallacy of Euler’s con- jecture [1][5]:
Consider 6 regiments and from each regiment take 6 officers of 6 different ranks. Hence, we have a total of 36 officers.
Question (Euler): Is it possible to arrange these 36 officers in the form of a 6×6 matrix satisfying the following condition?
Each row and each column of the 6×6 matrix contains only one officer from each rank and only one officer from each column.
Euler conjectured that such an arrangement is impossible.
Representation: Let us denote the ranks and regiments by the integers 1,2,3,4,5,6. Then associate to each officer the ordered pair (i, j) with 1≤i, j ≤n, where the first co-ordinateirepresents the officer’s rank and the second co-ordinate j, the officer’s regiment.
Then Euler’s conjecture is equivalent to constructing a 6×6 matrix with distinct entries (i, j) with 1 ≤ i, j ≤ n, in such a way that each row has all of the six numbers 1,2,3,4,5,6 in some order as the first co-ordinate and each column has all of the six numbers 1,2,3,4,5,6 in some order as the second co-ordinate.
Euler’s 36 officers problem was verified by Tarry [5] by system- atic enumeration.
Euler’s conjecture: Euler further conjectured that such an ar- rangement/matrix is impossible for all square matrices of order (4n+ 2)×(4n+ 2) for all n ≥ 2. This research earned the trio, Bose, Shrikhande, and Parker, the sobriquet Euler’s spoiler, and created such a stir that even the New York Times brought the news out on its first page.
But Bose, Shrikhande, and Parker proved the existence of such a matrix for eachn ≥2. The following remarkable 10×10 matrix is a counterexample to Euler’s conjecture forn = 2.
For convenience, let us usea, b, c, d, e, f, g, h, i, j for the first co- ordinate instead of 1,2,3,4,5,6,7,8,9,10 and A, B, C, D, E, F, G, H, I, Jfor the second co-ordinate in place of 1,2,3,4,5,6,7,8,9,10.
For instance,gI denotes an officer of the rank 7 in the regiment 9
2.2. ELEMENTARY COUNTING PRINCIPLES 51 and aJ an officer of rank 1 of the regiment 10.
M =
1 2 3 4 5 6 7 8 9 10
1 jJ f G eH dI iA hC gE aB bD cF
2 gF aA jG f H eI iB hD bC cE dJ
3 hE gJ bB aG jH f I iC cD dF eA
4 iD hF gA cC bG aH jI dE eJ f B
5 aI iE hJ gB dD cG bH eF f A jC
6 cH bI iF hA gC eE dG f J jB aD
7 eG dH cI iJ hB gD f F jA aC bE
8 bA cB dC eD f E jF aJ gG hH iI
9 dB eC f D jE aF bJ cA hI iG gH
10 f C jD aE bF cJ dA eB iH gI hG
Each row and each column of the matrix M has all the letters a, b, . . . , j as first co-ordinate in some order. Similarly, each row and each column of the matrix M has all the letters A, B, . . . , J in some order as the second co-ordinate.
2.2 Elementary Counting Principles
First we recall the following elementary “sum rule” concerning sets. We denote by|A|, the number of elements in the set A.
Fact 2.2.0.1 (Sum rule):
Consider two disjoint finite sets A and B, that is, A and B are finite sets with A∩B =∅. The number of elements in the union
|A∪B| is |A|+|B|.Symbolically,
|A∪B|=|A|+|B|.
There is nothing sacred about the number two in the above fact. More generally, ifA1, A2,· · · , An are n finite pairwise mutu- ally disjoint sets (that is, Ai∩Aj =∅ fori, j with 1≤i < j ≤n) then the number of elements in the union A1 ∪A2 ∪ · · · ∪An is given by the equation
|A1∪A2∪ · · · ∪An|=|A1|+|A2|+· · ·+|An|.
The whole is equal to the sum of its parts.
Let us recall the idea of the Cartesian product of two or more given sets.
If A and B are two given sets then the Cartesian product or simply product ofAandB is denoted byA×B and defined as the set of all ordered pairs (a, b) with a∈A and b∈B. Symbolically,
A×B ={(a, b)|a∈A, b∈B}
More generally, the product ofnsetsA1, A2,· · · , Anis denoted by A1×A2× · · · ×An and is defined as the set of all orderedn-tuples (a1, a2,· · · , an) withai ∈Ai fori= 1,2,· · · , n. ai is called theith component of the n-tuple (a1, a2,· · · , an)
We declare the n-tuple (a1, a2,· · · , an) to be equal to the n- tuple (a′1, a′2,· · · , a′n) if and only if ai =a′i for all i= 1,2,· · · , n.
Example 2.2.1 (Cartesian product):
Consider two sets A and B whereA ={1,2,3} and B ={2,4}. Then the product of A and B is
A×B ={(1,2),(1,4),(2,2),(2,4),(3,2),(3,4)} Its geometric representation is shown in Figure 2.1:
Figure 2.1: An example illustrating a Cartesian product
2.2. ELEMENTARY COUNTING PRINCIPLES 53 Example 2.2.2 (Product of two intervals):
Consider the two-dimensional Cartesian plane xy. Consider two intervals [1,3] = {x | x real with 1 ≤ x ≤ 3} in the x axis and [2,3]{y|y real with 2≤y≤3}in theyaxis. Then the geometric interpretation of the product [1,3]×[2,3] is the set of all points of the rectangle ABCD in the xy plane where the co-ordinates of A,B,C,D are respectively (1,2),(1,3),(3,2),(3,3) (see Figure 2.2).
Figure 2.2: An example illustrating product of intervals
Example 2.2.3 (Right circular cylinder as Cartesian product): The product of a circle in thexyplane and a closed interval in the z axis is a right circular cylinder in the three dimensional space R3.