The Fundamental Theorem of Enumeration,
independently discovered by several anonymous
cave-dwellers, states that
| A| =
∑1
a∈ A
Doron Zeilberger
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Contents
1. Introduction
2. Permutations
3. Generating Functions
4. Binomials
5. Inclusion-Exclusion
6. Möbius Inversion
7. Cauchy-Frobenius Theorem
8. Pólya-Redfield Theorem
Bibliography
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We are interested in counting objects
But what is an answer?
Example 1. How many even numbers are there less than
n? The answer is in explicit and compact form: the integer part of n/2, i.e., [ n/2].
Example 2. How many permutations α of {1, 2, . . . , n}
are there so that α(i ) 6= i for all i.
The answer is explicit but involves a summation depending on n:
n
(−1)i
.
dn = n! ∑
i!
i =0
Sometimes explicit answers cannot be found.
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Problem settings
Count the number of objects
• without symmetry: How many prime numbers are
there ≤ n?
• with symmetry: How many cubes are there with
green/blue faces?
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Problem
Consider red and blue beads. Beads of the same colour
are indistinguishable.
How many necklaces are there of n beads?
The solution depends on what is considered to be a
necklace.
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Four beads in line
2n = 16 ways to put n beads in line.
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10 open necklaces
These can be turned over.
This means that some of the previous designs give the
same open necklace.
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6 (closed) necklaces
These can be rotated and turned over.
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4 patterns of necklaces
We allow change • and •.
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Cayley 1875: Alkanes Cn H2n+2
How many isomers Cn H2n+2 are there?
Hexane and isohexane: both C6 H14 but
their structures are quite different.
Counting problems can be difficult
because of symmetries.
Rains and Sloane (1999) →
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
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count
1
1
1
2
3
5
9
18
35
75
159
355
802
1858
4347
C8 H18
... and they have all names!
Octane
3-Methylheptane
2,2-Dimethylhexane
2,4-Dimethylhexane
3,3-Dimethylhexane
3-Ethylhexane
2,2,4-Trimethylpentane
2,3,4-Trimethylpentane
3-Ethyl-3-methylpentane
2-Methylheptane
4-Methylheptane
2,3-Dimethylhexane
2,5-Dimethylhexane
3,4-Dimethylhexane
2,2,3-Trimethylpentane
2,3,3-Trimethylpentane
3-Ethyl-2-methylpentane
2,2,3,3-Tetramethylbutane
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Notation
• N = {0, 1, . . . }, Z, Q, R and C
• We sometimes consider the field (R, +, ·) of real
numbers, but what is needed is an arbitrary field
(sometimes of characteristic 0).
• [1, n] = {1, 2, . . . , n}.
• The Kronecker symbol:
• k | n:
δnk =
k divides n for k ∈ [1, n].
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(
1
if n = k
0
if n 6= k
• For a finite set X, let | X | denote its size.
A set X is an n-set (n-subset of Y) if | X | = n (X ⊆
Y).
• 2X denotes the power set of X.
• n! = 1 · 2 · . . . · n is the factorial of a nonnegative
integer n with 0! = 1.
• For 0 ≤ k ≤ n from N,
n
n!
=
k
k!(n − k )!
is the binomial coefficient “n choose k”.
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• S A is the set of all permutations of the set A.
• In summations such as
∑
nz + k
k≤ẑ ≤n
the summation is over the elements z.
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Introduction to combinatorial counting
A typical technique is double counting, where the number of objects is counted in two different ways.
Lemma (Handshaking). Let A = { A1, A2, . . . , An } be a
family of subsets Ai of a finite set X. The degree of an element
x ∈ X is defined by
d( x ) = |{i | x ∈ Ai }|.
Then
n
∑ d ( x ) = ∑ | Ai | .
x∈X
i =1
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Proof
Let X = { x1, x2, . . . , xm }, and consider the matrix M =
(mij )n×m , where
(
1 if xi ∈ A j ,
mij =
0 otherwise.
Then d( xi ) equals the number of ones in the ith row,
and | A j | equals the number one ones in the jth column.
Hence the claim.
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Example: X = [1, 4]
A1 = {1, 3, 4}, A2 = {1, 2, 4}, A3 = {2, 3}.
Then


1 1 0


0 1 1
M=

1 0 1
1 1 0
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Changing summation order
In this example, double counting happens in the change
of summation order. Let f : N → N be a function of
nonnegative integers, then
n
i
n
n
∑ ∑ f (i, k) = ∑ ∑ f (i, k) .
i =1 k =1
k =1 i = k
Indeed, on both sides, f (i, k ) is counted once for each
pair (i, k ) with 1 ≤ k ≤ i ≤ n.
One could write the summation as
∑
f (i, k )
1≤k̂ ≤iˆ≤n
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A geometric proof of n2 = ∑nk=1 2k − 1
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Pigeon hole principle
Two sets have the same size if and only if there is a
bijection between them.
Lemma (Pigeon Holes). Let f : X → Y be a function. If
|Y | < | X |, then f is not injective.
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Pigeon hole: Chinese remainder thm
Let n, m ∈ N be relatively prime and 0 ≤ a < n and
0 ≤ b < m. Then there exists an integer x such that
x ≡ a (mod n) and x ≡ b (mod m ).
Proof. Holes are the residue classes modulo m.
Pigeons are a + in for i = 0, 1, . . . , m − 1.
Suppose a + in ≡ a + jn (mod m ) for some i < j. Then
m | a + jn − a − in = ( j − i )n, and so m | j − i. But j − i < m;
a contradiction.
Hence a, a + n, . . . , a + (m − 1)n are distinct modulo m,
and so b ≡ a + in (mod m ) for some i < m.
Hence a + in = bm + x, and x is as required.
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