The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5

·

4

·

3 = 60

The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5

·

4

·

3 = 60

The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5

·

4

·

3 = 60

The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5 ·

4

·

3 = 60

The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5 · 4 ·

3 = 60

The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5 · 4 · 3

= 60

The Principle of Overcounting

In combinatorics, we often have to count in a two-step procedure:

1 Use a “wrong” procedure to count, which yields a numberthat’s too high.

2 Then figure out how wrong our answer was, and reduce it tothe correct number.

Example

How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:

5 · 4 · 3 = 60

The Principle of Overcounting

ABC

ABD

ABE

ACB

ACD

ACE

ADB

ADC

ADE

AEB

AEC

AED

BAC

BAD

BAE

BCA

BCD

BCE

BDA

BDC

BDE

BEA

BEC

BED

CAB

CAD

CAE

CBA

CBD

CBE

CDA

CDB

CDE

CEA

CEB

CED

DAB

DAC

DAE

DBA

DBC

DBE

DCA

DCB

DCE

DEA

DEB

DEC

EAB

EAC

EAD

EBA

EBC

EBD

ECA

ECB

ECD

EDA

EDB

EDC

The Principle of Overcounting

The Principle of Overcounting

ABC

ABD

ABE

ACB

ACD

ACE

ADB

ADC

ADE

AEB

AEC

AED

BAC

BAD

BAE

BCA

BCD

BCE

BDA

BDC

BDE

BEA

BEC

BED

CAB

CAD

CAE

CBA

CBD

CBE

CDA

CDB

CDE

CEA

CEB

CED

DAB

DAC

DAE

DBA

DBC

DBE

DCA

DCB

DCE

DEA

DEB

DEC

EAB

EAC

EAD

EBA

EBC

EBD

ECA

ECB

ECD

EDA

EDB

EDC

{A,B,C}{A,B,D}{A,B,E}{A,C ,D}{A,C ,E}{A,D,E}{B,C ,D}{B,C ,E}{B,D,E}{C ,D,E}

The Principle of Overcounting

ABC

ABD

ABE

ACB

ACD

ACE

ADB

ADC

ADE

AEB

AEC

AED

BAC

BAD

BAE

BCA

BCD

BCE

BDA

BDC

BDE

BEA

BEC

BED

CAB

CAD

CAE

CBA

CBD

CBE

CDA

CDB

CDE

CEA

CEB

CED

DAB

DAC

DAE

DBA

DBC

DBE

DCA

DCB

DCE

DEA

DEB

DEC

EAB

EAC

EAD

EBA

EBC

EBD

ECA

ECB

ECD

EDA

EDB

EDC

{A,B,C}{A,B,D}{A,B,E}{A,C ,D}{A,C ,E}{A,D,E}{B,C ,D}{B,C ,E}{B,D,E}{C ,D,E}

60

6= 10 =

(53

)

Another Example

Another Example

How many ways to seat three people around a round table?

Initial (wrong) answer: 3! = 6.

Another Example

Another Example

How many ways to seat three people around a round table?

3

Initial (wrong) answer: 3! = 6.

Another Example

Another Example

How many ways to seat three people around a round table?

3

2

Initial (wrong) answer: 3! = 6.

Another Example

Another Example

How many ways to seat three people around a round table?

3

21

Initial (wrong) answer: 3! = 6.

Another Example

Another Example

How many ways to seat three people around a round table?

3

21

Initial (wrong) answer: 3! = 6.

Another Example

Another Example

Another Example

Another Example

Correct answer:3!

3= 2.

A More Complicated Example

Example

You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?

Initial answer: 2 · 2 · 2 = 8.

A More Complicated Example

Example

You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?

2

Initial answer: 2 · 2 · 2 = 8.

A More Complicated Example

Example

You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?

2

2

Initial answer: 2 · 2 · 2 = 8.

A More Complicated Example

Example

You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?

2

2

2

Initial answer: 2 · 2 · 2 = 8.

A More Complicated Example

Example

You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?

2

2

2

Initial answer: 2 · 2 · 2 = 8.

A More Complicated Example

Example

You have two colors, red and blue. How many ways are there topaint the corners of an equilateral triangle?

Initial (wrong) answer: 2 · 2 · 2 = 8.

A More Complicated Example

A More Complicated Example

A More Complicated Example

A More Complicated Example

Correct answer: 4.

An Even More Complicated Example

An Even More Complicated Example

You have two colors, red and blue. How many ways are there tocolor the corners of a square?

An Even More Complicated Example

An Even More Complicated Example

You have two colors, red and blue. How many ways are there tocolor the corners of a square?

22

2 2

Wrong answer: 24 = 16

An Even More Complicated Example

An Even More Complicated Example

An Even More Complicated Example

An Even More Complicated Example

Correct answer: 6.

Absurdly Complicated Example

Absurdly Complicated Example

How many ways are there to paint the corners of a square withfour colors? Count colorings as the same if one can be rotated orflipped to look like the other.

The initial (wrong) answer: 44 = 256.

I don’t want to draw all those out,let alone figure out where the repeats are.

We need a faster way to predict the answer, withoutexamining all the possibilities by hand.

The answer: group actions!

R90

R90

Observations

In our example:

Every coloring we wanted was in the whole list of 24

possibilities.

Some of those were equivalent and shouldn’t be countedseparately.

We considered two colorings equivalent when one could betransformed into the other by a rotation.

That is, we were looking at the orbits of the colorings underthe action of the group of rotations {R0,R90,R180,R270}.Our problem: Counting the orbits.

The Set-Up

G acts on S where

S = all coloringsG = symmetry group of the figure

“distinguishable” = in different orbits

Combinatorics Strategy

1 Think of all possible ways (including duplicates).

2 Decide what makes two ways count as equivalent.

3 Decide what group action has those equivalence classes as itsorbits.

4 Use the not-Burnside Theorem to count the orbits.

Combinatorics Strategy

1 Think of all possible ways (including duplicates).

2 Decide what makes two ways count as equivalent.

3 Decide what group action has those equivalence classes as itsorbits.

4 Use the not-Burnside Theorem to count the orbits.

Combinatorics Strategy

1 Think of all possible ways (including duplicates).

2 Decide what makes two ways count as equivalent.

3 Decide what group action has those equivalence classes as itsorbits.

4 Use the not-Burnside Theorem to count the orbits.

Combinatorics Strategy

1 Think of all possible ways (including duplicates).

2 Decide what makes two ways count as equivalent.

3 Decide what group action has those equivalence classes as itsorbits.

4 Use the not-Burnside Theorem to count the orbits.

The not-Burnside Theorem

Theorem (not-Burnside)

Let G act on S. For each g ∈ G, let ψ(g) = |Fix(g)|. Then thenumber of orbits of S under the action of G is

1

|G |∑g∈G

ψ(g).

Proof Sketch:

Count {(g , s) ∈ G × S : s · g = s} in two different ways. �

The not-Burnside Theorem

Theorem (not-Burnside)

Let G act on S. For each g ∈ G, let ψ(g) = |Fix(g)|. Then thenumber of orbits of S under the action of G is

1

|G |∑g∈G

ψ(g).

Proof Sketch:

Count {(g , s) ∈ G × S : s · g = s} in two different ways. �

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0

16

R90

2

R180

4

R270

224

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =

1

4· 24 = 6.

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0 16R90

2

R180

4

R270

224

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =

1

4· 24 = 6.

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0 16R90 2R180

4

R270

224

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =

1

4· 24 = 6.

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0 16R90 2R180 4R270

224

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =

1

4· 24 = 6.

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0 16R90 2R180 4R270 2

24

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =

1

4· 24 = 6.

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0 16R90 2R180 4R270 2

24

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =

1

4· 24 = 6.

Applying not-Burnside

Even More Complicated Example, revisited

You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.

Our symmetry group is {R0,R90,R180,R270} ≤ D8.

g ψ(g)

R0 16R90 2R180 4R270 2

24

Thus the number of distinguishable colorings is

1

|G |∑g∈G

ψ(g) =1

4· 24 = 6.

Applying not-Burnside

Absurdly Complicated Example, revisited

How many ways are there to paint the corners of a square withfour colors? Count colorings as the same if one can be rotated orflipped to look like the other.

Note that our symmetry group is D8.

Applying not-Burnside

g ψ(g)

R0

256

R90

4

R180

16

R270

4

H

16

V

16

D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90

4

R180

16

R270

4

H

16

V

16

D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180

16

R270

4

H

16

V

16

D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270

4

H

16

V

16

D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H

16

V

16

D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H 16V

16

D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H 16V 16D1

64

D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H 16V 16D1 64D2

64440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H 16V 16D1 64D2 64

440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H 16V 16D1 64D2 64

440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =

1

8· 440 = 55.

Applying not-Burnside

g ψ(g)

R0 256R90 4R180 16R270 4H 16V 16D1 64D2 64

440

Thus the number of distinguishable colorings is

1

|D8|∑g∈D8

ψ(g) =1

8· 440 = 55.