Introduction to Combinatorics: Basic Counting Techniques - pjwstk

Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction to Combinatorics:
Basic Counting Techniques
Introduction
Basic
Counting
General
Techniques
Marcin Sydow
Project co-nanced by European Union within the framework of European Social Fund

Literature
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Combinatorics:
D.Knuth et al. Concrete Mathematics (also available in
Polish, PWN 1998)
M.Libura et al. Wykªady z Matematyki Dyskretnej, cz.1
Kombinatoryka, skrypt WSISIZ, Warszawa 2001
M.Lipski Kombinatoryka dla programistów, WNT 2004
Van Lint et al. A Course in Combinatorics, Cambridge
2001
Graphs:
R.Wilson Introduction to Graph Theory (also available in
Polish, PWN 2000)
R.Diestel Graph Theory, Springer 2000

Contents
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Basic Counting
Reminder of basic facts
Binomial Coecient
Multinomial Coecient
Multisets
Set Partitions
Number Partitions
Permutations and Cycles
General Techniques
Pigeonhole Principle
Inclusion-Exclusion Principle
Generating Functions

General Basic Ideas for Counting
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
create easy-to-count representations of counted objects
product rule: multiply when choices are independent
sum rule: sum up exclusive alternatives
combinatorial interpretation proving technique
These ideas can be used not only to count objects but also to
easily prove non-trivial discrete-maths identities (examples
soon)

Counting Basic Objects
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
Denotations: # means: the number of
[n] means: the set {1,...,n}, for n ∈ N
n, m ∈ N
(# functions from [m] into [n]) |Fun([m], [n])| =
General
Techniques

Counting Basic Objects
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
Denotations: # means: the number of
[n] means: the set {1,...,n}, for n ∈ N
n, m ∈ N
(# functions from [m] into [n]) |Fun([m], [n])| = n m
(# subsets of an n-element set) |P([n])| =
General
Techniques

Counting Basic Objects
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Denotations: # means: the number of
[n] means: the set {1,...,n}, for n ∈ N
n, m ∈ N
(# functions from [m] into [n]) |Fun([m], [n])| = n m
(# subsets of an n-element set) |P([n])| = 2 n
(# injections as above) |Inj([m], [n])| =

Counting Basic Objects
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Denotations: # means: the number of
[n] means: the set {1,...,n}, for n ∈ N
n, m ∈ N
(# functions from [m] into [n]) |Fun([m], [n])| = n m
(# subsets of an n-element set) |P([n])| = 2 n
(# injections as above) |Inj([m], [n])| = n m for n ≥ m (called
falling factorial)
# ordered placings of m dierent balls into n dierent boxes :

Counting Basic Objects
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Denotations: # means: the number of
[n] means: the set {1,...,n}, for n ∈ N
n, m ∈ N
(# functions from [m] into [n]) |Fun([m], [n])| = n m
(# subsets of an n-element set) |P([n])| = 2 n
(# injections as above) |Inj([m], [n])| = n m for n ≥ m (called
falling factorial)
# ordered placings of m dierent balls into n dierent boxes : n ©m
(called increasing power)
(# permutations of [n]) |S n | =

Counting Basic Objects
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Denotations: # means: the number of
[n] means: the set {1,...,n}, for n ∈ N
n, m ∈ N
(# functions from [m] into [n]) |Fun([m], [n])| = n m
(# subsets of an n-element set) |P([n])| = 2 n
(# injections as above) |Inj([m], [n])| = n m for n ≥ m (called
falling factorial)
# ordered placings of m dierent balls into n dierent boxes : n ©m
(called increasing power)
(# permutations of [n]) |S n | = n!

Binomial Coecient
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
# k-subsets of [n]
( n
k
)
= nk
k!
N ∋ k ≤ n ∈ N
(there is also possible a more general formulation for
non-natural numbers)

Recursive formulation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
( n
k
)
=
( n − 1
k − 1
) ( n − 1
+
k
(also known as Pascal's triangle)
)
General
Techniques

Recursive formulation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
( n
k
)
=
( n − 1
k − 1
) ( n − 1
+
k
(also known as Pascal's triangle)
how to prove it without (almost) any algebra
(use combinatorial interpretation idea)
)

Recursive formulation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
( n
k
)
=
( n − 1
k − 1
) ( n − 1
+
k
(also known as Pascal's triangle)
how to prove it without (almost) any algebra
(use combinatorial interpretation idea)
(consider whether one xed element of [n] belongs to the
k-subset or not)
)

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Symmetry:
( n
k
)
=
( n
n − k
)
Introduction
Basic
Counting
General
Techniques

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
( ) ( )
n n
Symmetry: =
k n − k
(each k-subset denes (n-k)-subset its complement)
Basic
Counting
General
Techniques

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
( ) ( )
n n
Symmetry: =
k n − k
(each k-subset denes (n-k)-subset its complement)
n∑
( ) n
= 2 n
k
k=0

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
( ) ( )
n n
Symmetry: =
k n − k
(each k-subset denes (n-k)-subset its complement)
n∑
( ) n
= 2 n
k
k=0
(summing subsets of [n] by their multiplicity)

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
( n
k
) ( k
m
) ( n
=
m
) ( n − m
n − k
)
Introduction
Basic
Counting
General
Techniques

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
( ) ( ) ( ) ( )
n k n n − m
=
k m m n − k
(consider m-element subset of o k-element subset of [n])
General
Techniques

Basic properties
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
( ) ( ) ( ) ( )
n k n n − m
=
k m m n − k
(consider m-element subset of o k-element subset of [n])
( m + n
k
)
=
k∑
( m
s
s=0
) ( n
k − s
)

Binomial Theorem
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(x + y) n =
n∑
( ) n
x k y n−k
k
k=0
(explanation: each term on the right may be represented as a
k-subset of [n])
Corollaries:
# odd subsets of [n] equals # even subsets

Binomial Theorem
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(x + y) n =
n∑
( ) n
x k y n−k
k
k=0
(explanation: each term on the right may be represented as a
k-subset of [n])
Corollaries:
# odd subsets of [n] equals # even subsets (substitute
x=1 and y=-1)

Binomial Theorem
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(x + y) n =
n∑
( ) n
x k y n−k
k
k=0
(explanation: each term on the right may be represented as a
k-subset of [n])
Corollaries:
# odd subsets of [n] equals # even subsets (substitute
x=1 and y=-1)
n∑
( ) n
k = n2 n−1
k
k=0
(take derivative (as function of x) of both sides and then
subsitute x=y=1)

Multinomial Coecient
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(a generalisation of the binomial coecient)
# colourings of n balls with at most m dierent colours so that
exactly k i ∈ N balls have the i-th colour
(
)
n
=
k 1 k 2 . . . k m
n!
k 1 ! · . . . · k m !

Multinomial Coecient
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(a generalisation of the binomial coecient)
# colourings of n balls with at most m dierent colours so that
exactly k i ∈ N balls have the i-th colour
(
)
n
=
k 1 k 2 . . . k m
n!
k 1 ! · . . . · k m !
(x the sequence (k) i and x a permutation of [n] to be
coloured)

Multinomial Theorem
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
(x 1 +. . .+x m ) n =
∑
k 1 , . . . , k m ∈ N
k 1 + . . . + k m = n
(
)
n
x k 1 km
1 ·. . .·xm
k 1 k 2 . . . k m
Basic
Counting
General
Techniques
Corollary:
∑
k 1 , . . . , k m ∈ N
k 1 + . . . + k m = n
(
n
k 1 k 2 . . . k m
)
= m n

Multinomial Theorem
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
(x 1 +. . .+x m ) n =
∑
k 1 , . . . , k m ∈ N
k 1 + . . . + k m = n
(
)
n
x k 1 km
1 ·. . .·xm
k 1 k 2 . . . k m
Basic
Counting
General
Techniques
Corollary:
∑
k 1 , . . . , k m ∈ N
k 1 + . . . + k m = n
(
)
n
= m n
k 1 k 2 . . . k m
(substitute all 1's on the left-hand side)

Recursive Formula
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(
) (
n
=
k 1 k 2 . . . k m
) (
n − 1
+
k 1 − 1 k 2 . . . k m
(
+ . . . +
n − 1
k 1 k 2 . . . k m − 1
)
n − 1
+
k 1 k 2 − 1 . . . k m
)

Recursive Formula
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(
) (
n
=
k 1 k 2 . . . k m
) (
n − 1
+
k 1 − 1 k 2 . . . k m
(
+ . . . +
n − 1
k 1 k 2 . . . k m − 1
)
n − 1
+
k 1 k 2 − 1 . . . k m
(explanation: x one element of [n] and analogously to the
binomial theorem)
)

Multisets
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Element of type x i has k i (identical) copies.
Q =< k 1 ∗ x 1 , . . . , k n ∗ x n >
Introduction
Basic
Counting
General
Techniques
Subset S of Q:
S =< m 1 ∗ x 1 , . . . , m n ∗ x n >
(0 ≤ m i ≤ k i , for i ∈ [n])
Fact:
# subsets of Q =
|Q| = k 1 + . . . + k n

Multisets
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Element of type x i has k i (identical) copies.
Q =< k 1 ∗ x 1 , . . . , k n ∗ x n >
Introduction
Basic
Counting
General
Techniques
Subset S of Q:
S =< m 1 ∗ x 1 , . . . , m n ∗ x n >
(0 ≤ m i ≤ k i , for i ∈ [n])
|Q| = k 1 + . . . + k n
Fact:
# subsets of Q = (1 + k 1 ) · (1 + k 2 ) · . . . · (1 + k n )

Partitions of number n into sum of k terms
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
P(n, k)
# k-partitions of n, n ∈ N
Partition of n: n = a 1 + . . . + a k , a 1 ≥ . . . ≥ a k > 0
Recursive formula:
P(n, k) = P(n − 1, k − 1) + P(n − k, k)

Partitions of number n into sum of k terms
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
P(n, k)
# k-partitions of n, n ∈ N
Partition of n: n = a 1 + . . . + a k , a 1 ≥ . . . ≥ a k > 0
Recursive formula:
P(n, k) = P(n − 1, k − 1) + P(n − k, k)
(either a k = 1 or all blocks can be decreased by 1 element)

Partitions of number n into sum of k terms
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
P(n, k)
# k-partitions of n, n ∈ N
Partition of n: n = a 1 + . . . + a k , a 1 ≥ . . . ≥ a k > 0
Recursive formula:
P(n, k) = P(n − 1, k − 1) + P(n − k, k)
(either a k = 1 or all blocks can be decreased by 1 element)
Fact:
P(n, k) =
k∑
P(n − k, i)
i=0

Partitions of number n into sum of k terms
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
# k-partitions of n, n ∈ N
P(n, k)
Partition of n: n = a 1 + . . . + a k , a 1 ≥ . . . ≥ a k > 0
Introduction
Basic
Counting
General
Techniques
Recursive formula:
P(n, k) = P(n − 1, k − 1) + P(n − k, k)
(either a k = 1 or all blocks can be decreased by 1 element)
Fact:
k∑
P(n, k) = P(n − k, i)
i=0
(decrease by 1 each of i terms greater than 1)

Set Partitions
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Partition of nite set X into k blocks: Π k = A 1 , . . . , A k so that:
∀ 1≤i≤k A i ≠ ∅
A 1 ∪ . . . ∪ A k = X
∀ 1≤i

Stirling Number of the 2nd kind
Introduction
to Combinatorics:
Basic
Counting
Techniques
{ n
k
}
= |Π k ([n])|
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(# partitions of [n] into k blocks)
Recursive formula:
{ } { }
n n
= 1 = 0 for n > 0
n 0
{ n
k
}
=
{ n − 1
k − 1
} { n − 1
+ k
k
}

Stirling Number of the 2nd kind
Introduction
to Combinatorics:
Basic
Counting
Techniques
{ n
k
}
= |Π k ([n])|
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
(# partitions of [n] into k blocks)
Recursive formula:
{ } { }
n n
= 1 = 0 for n > 0
n 0
{ n
k
}
=
{ n − 1
k − 1
} { n − 1
+ k
k
(a xed element constitutes a singleton-block or belongs to one
of bigger k blocks)
}

Equivalence Relations and Surjections
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
Bell Number
B n =
n∑
{ n
k
k=0
(# equivalence relations on [n])
}
General
Techniques
# surjections from [m] onto [n], m ≥ n =

Equivalence Relations and Surjections
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
Bell Number
B n =
n∑
{ n
k
k=0
(# equivalence relations on [n])
}
General
Techniques
# surjections from [m]
{
onto
}
[n], m ≥ n =
m
|Sur([m], [n])| = m! ·
n

Relation between x n , x n , x n (with Stirling 2-nd order numbers)
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
n∑
{ n
x n =
k
k=0
}
· x n =
n∑
{ n
(−1) n−k ·
k
k=0
}
· x n

Permutations
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Inj([n], [n])
Permutations on [n] constitute a group denoted by S n
composition of 2 permutations gives a permutation on [n]
identity permutation is a neutral element (e)
inverse of permutation f is a permutation f −1 on [n]

Examples
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Inverse:
( ) 12345
f =
53214
( ) 12345
g =
25314
f −1 =
( 12345
43251
The group is not commutative: fg is dierent than gf:
( ) 12345
fg =
34251
gf =
( 12345
43521
)
)

Decomposition into Cycles
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
A cycle is a special kind of permutation.
Each permutation can be decomposed into disjoint cycles:
Example:
decomposition is unique
the cycles are commutative
f =
( 12345
53214
)
f = f ′ f ′′ = [1, 5, 4][2, 3] = [2, 3][1, 5, 4] = f ′′ f ′ = f

Stirling Number of the 1st kind
Introduction
to Combinatorics:
Basic
Counting
Techniques
# permutations consisting of exactly k cycles:
[ n
k
]
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
n∑
[ ] n
= n!
k
k=0
Recursive Formula:
[ ] [ n n − 1
=
k k − 1
] [ n − 1
+ (n − 1)
k
]

Stirling Number of the 1st kind
Introduction
to Combinatorics:
Basic
Counting
Techniques
# permutations consisting of exactly k cycles:
[ n
k
]
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
n∑
[ ] n
= n!
k
k=0
Recursive Formula:
[ ] [ n n − 1
=
k k − 1
] [ n − 1
+ (n − 1)
k
(x a single element: it either itself constitutes a 1-cycle or can
be at one of the n-1 positions in the k cycles)
]

Expressing decreasing and increasing powers in terms
of normal powers with the help of Stirling Numbers of the 1-st kind
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
x n =
n∑
[ n
k
k=0
x n =
k=0
]
(−1) n−k x k
n∑
[ ] n
k
x k

(Reminder of the opposite)
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
n∑
{ n
x n =
k
k=0
}
· x n =
n∑
{ n
(−1) n−k ·
k
k=0
}
· x n

Other (amazing) connections between Stirling
Numbers of both kinds
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
n∑
[ ] { }
n k
(−1) n−k = [m == n]
k m
k=0
n∑
{ } [ ] n k
(−1) n−k = [m == n]
k m
k=0

Type of permutation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Permutation f ∈ S n has type (λ 1 , . . . , λ n ) i its decomposition
into disjoint cycles contains exactly λ i cycles of length i.
Example:
f =
f = [1, 7, 6, 3], [2, 5], [4], [8, 9]
( 123456789
751423698
)
Thus, the type of f is (1, 2, 0, 1, 0, 0, 0, 0, 0).
We equivalently denote it as: 1 1 2 2 4 1

Inversion in a permutation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
Inversion in a permutation f = (a 1 , . . . , a n ) ∈ S n is a pair
(a i , a j ) so that i < j ≤ n and a i > a j
The number of inversions in permutation f is denoted as I (f )
What is the minimum/maximum value of I (f )
How does it relate to sorting
How to eciently compute I (f )

Transpositions
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
A permutation that is a cycle of length 2 is called transposition
Fact: Each permutation f is a composition of exactly I (f )
transpositions of neighbouring elements

Sign of Permutation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
assume, f ∈ S n :
(denition)
sgn(f ) = (−1) I (f )
Introduction
Basic
Counting
General
Techniques
sgn(fg) = sgn(f ) · sgn(g)
sgn(f −1 ) = sgn(f )
We say that a permutation is even when sgn(f ) = 1 and odd
otherwise
Fact: sgn(f ) = (−1) k−1 for any f being a k-cycle

Computing the sign of a permutation
Introduction
to Combinatorics:
Basic
Counting
Techniques
Marcin
Sydow
Introduction
Basic
Counting
General
Techniques
For any permutation f ∈ S n of the type (1 λ 1
. . . n λn ) its sign
can be computed as follows:
∑ ⌊n/2⌋
sgn(f ) = (−1) j=1 λ 2j
(only even cycles contribute to the sign of the permutation)

General Techniques
Introduction
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Marcin
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Introduction
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Pigeonhole Principle
Inclusion-Exclusion Principle
Generating Functions

Pigeonhole Principle (Pol. Zasada szuadkowa)
Introduction
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Marcin
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Introduction
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Let X,Y be nite sets, f ∈ Fun(X , Y ) and |X | > r · |Y | for some
r ∈ R + . Then, for at least one y ∈ Y , |f −1 ({y})| > r.
(or equivalently: if you put m balls into n boxes then at least
one box contains not less than m/n balls)

Examples
Introduction
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Marcin
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Introduction
Any 10-subset of [50] contains two dierent 5-subsets that have
the same sum of elements.
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Examples
Introduction
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Marcin
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Introduction
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Any 10-subset of [50] contains two dierent 5-subsets that have
the same sum of elements.
(Hair strands theorem, etc.):
At the moment, there exist two people on the earth that have
exactly the same number of hair strands

Inclusion-Exclusion Principle
(Pol. Zasada Wª¡cze«-Wyª¡cze«)
Introduction
to Combinatorics:
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Marcin
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Introduction
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For any non-empty family A = {A 1 , . . . , A n } of subsets of a
nite set X, the following holds:
|
n⋃
A i | =
i=1
n∑
(−1) i−1
i=1
(proof by induction on n)
∑
1≤p 1

Example: number of Derangements
Introduction
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Marcin
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Introduction
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A derangement (Pol. nieporz¡dek) is a permutation f ∈ S n so
that f (i) ≠ i, for i ∈ [n].
D n is the set of all derangements on [n]. |D n | is denoted as !n.
Theorem: !n = |D n | = n! ∑ n
i=0
(−1) i
i!

Proof of the formula for !n
Introduction
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Marcin
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Introduction
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Let A i = {f ∈ S n : f (i) = i}, for i ∈ [n]. Thus
!n = |S n | − |A 1 ∪ A 2 ∪ . . . ∪ A n | =
= n! −
n∑
(−1) i−1
i=1
∑
1≤p 1

Ratio of Derangements in Permutations
Introduction
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Marcin
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Introduction
Since !n = |D n | = n! ∑ n (−1) i
i=0 i!
The ratio of derangements: |D n |/|S n | while n → ∞ tends to 1 e
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e −1 =
∞∑
(−1) i 1 i! ≈ 0.368 . . . ,
i=0
(e ≈ 2.7182 . . . is the base of the natural logarithm)

Generating Functions
Introduction
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Marcin
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Introduction
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A generating function of an innite sequence a 0 , a 1 , . . . is a
power series:
A(z) =
where z is a complex variable
∞∑
a i z i
i=0
Generating functions is a powerful tool for representing,
manipulating and nding closed-form formulas for sequences
(especially recurrent sequences)

How to extract a sequence from its generating
function
Introduction
to Combinatorics:
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Marcin
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Introduction
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Let's view A(z) as a function of z, that is convergent in some
neighbourhood of z. Then we have:
a k = A(k) (0)
k!
(k-th factor in the Maclaurin series of A(z))

Examples
Introduction
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Marcin
Sydow
Introduction
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a i = [i = n] ∑ ∞
[i = n]z i
i=0
= z n
a i = c i ∑ ∞
i=0 ci z i = (1 − cz) −1 (geometric series)
a i = [m|i] ∑ ∞
z m·i
i=0
= 1
a i = (i!) −1 ∑ ∞
i=0
z i
i!
= e z
1−z m
(a) = (0, 1, 1/2, 1/3/, 1/4, . . .) ∑ ∞
i=1
z i
i
= −ln(1 − z)

Basic Operations
Introduction
to Combinatorics:
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Marcin
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Introduction
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Let A(z) and B(z) are the generating functions (GF) of
sequences (a i ) and (b i ), respectively, α ∈ R. Then:
GF of (a i + b i ) is A(z) + B(z) = ∑ ∞
i=0 (a i + b i )z i
GF of (α · a i ) is α · A(z) = ∑ ∞
i=0 α · a i · z i
GF of (a i−m ) is z m · A(z)

Further Examples
Introduction
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Marcin
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(0, 1, 0, 1 . . .) z(1 − z 2 ) −1
(1, 1/2, 1/3, . . .) −z −1 ln(1 − z)
a i = 1 (1 − z) −1
a i = (−1) i (1 + z) −1
i · a i z · A ′ (z)
a i = i z d dz (1 − z)−1 = z(1 − z) −2

Convolution of sequences
Introduction
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Marcin
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Introduction
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A convolution of two sequences (a i ) and (b i ) is a sequence c i :
c i =
i∑
a k · b i−k
k=0
and is denoted as: (c i ) = (a i ) ∗ (b i )
Convolution is commutative.
Fact:
∞∑
c i z i = A(z) · B(z)
i=0
(GF of (a i ) ∗ (b i ) is A(z) · B(z))

Example
Introduction
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Marcin
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Harmonic number:
Closed-form formula
H n =
n∑
i=1
GF for (H n ) is a convolution of (0, 1, 1/2, 1/3, . . .) and
(1, 1, 1, . . .). Thus, this GF is −(1 − z) −1 ln(1 − z).
1
i

Example: Closed-form formula Fibonacci Numbers
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Marcin
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thus (its GF is):
F i = F i−1 + F i−2 + [i = 1]
F (z) = zF (z) + z 2 F (z) + z
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z
F (z) =
1 − z − z 2
(1 − z − z 2 ) = (1 − az)(1 − bz), where a = (1 − √ 5)/2 and
b = (1 + √ 5)/2. Thus,
z
F (z) =
(1−az)(1−bz) = 1
(a−b) ( 1
(1−az) − 1
(1−bz) ) = ∑ ∞ a i −b i
i=0 a−b · z i .
Finally:
F i = √ 1 [( 1 + √ 5
) i − ( 1 − √ 5
) i ]
5 2
2

N-th order linear recurrent equations
Introduction
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Marcin
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a i = q(i) + q 1 · a i−1 + q 2 · a i−2 + . . . + q k · a i−k
where q(i) = a i , for i ∈ [k − 1] (initial conditions)
A(z) = A 0 (z) + q 1 · zA(z) + q 2 · z 2 A(z) + . . . + q k · z k A(z)
A(z) =
a 0 + a 1 z + . . . + a k−1 z k−1
1 − q 1 z − q 2 z 2 − . . . − q k z k

Introduction
to Combinatorics:
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Marcin
Sydow
Introduction
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Thank you for attention
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