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SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

Binomial Coefficients and Subsets Enumeration

the basis of combinatorics Master MOSIG : Mathematics for Computer Science Jean-Marc.Vincent@imag.fr September 2018 These notes are only the sketch of the lecture : the aim is to apply the basic counting techniques to the binomial coefficients and establish combinatorial equalities. References : Concrete Mathematics : A Foundation for Computer Science Ronald L. Graham, Donald E. Knuth and Oren Patashnik Addison-Wesley 1989 (chapter 5)

1 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

2 / 23 Binomial Coefficients and Subsets Enumeration

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SUBSET ENUMERATION

n

k

• is the number of ways to choose k elements among n elements

http://www-history.mcs.st-and.ac.uk/Biographies/Pascal.html

For all integers 0 k n n k

• = n(n − 1) · · · (n − k + 1)

k! (1)

3 / 23 Binomial Coefficients and Subsets Enumeration

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SUBSET ENUMERATION

n

k

• is the number of ways to choose k elements among n elements

http://www-history.mcs.st-and.ac.uk/Biographies/Pascal.html

For all integers 0 k n n k

• = n(n − 1) · · · (n − k + 1)

k! (1) Prove the equality by a combinatorial argument Hint : the number of sequences of k different elements among n is n(n − 1) · · · (n − k + 1) and the number of orderings of a set of size k is k!.

3 / 23 Binomial Coefficients and Subsets Enumeration

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BASIC PROPERTIES

n k

• =

n! k!(n − k)! (2) Prove it directly from Equation 1 For all integers 0 k n n k

• =
• n

n − k

• (3)

Prove it directly from 2 Prove it by a combinatorial argument

4 / 23 Binomial Coefficients and Subsets Enumeration

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BASIC PROPERTIES

n k

• =

n! k!(n − k)! (2) Prove it directly from Equation 1 For all integers 0 k n n k

• =
• n

n − k

• (3)

Prove it directly from 2 Prove it by a combinatorial argument Hint : bijection between the set of subsets of size k and ???. Exercise Give a combinatorial argument to prove that for all integers 0 k n : k n k

• = n

n − 1 k − 1

• (4)

4 / 23 Binomial Coefficients and Subsets Enumeration

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PASCAL’S TRIANGLE

Recurrence Equation The binomial coefficients satisfy n k

• =

n − 1 k − 1

• +

n − 1 k

• (5)

Prove it directly from Equation 1 Prove it by a combinatorial argument

5 / 23 Binomial Coefficients and Subsets Enumeration

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PASCAL’S TRIANGLE

Recurrence Equation The binomial coefficients satisfy n k

• =

n − 1 k − 1

• +

n − 1 k

• (5)

Prove it directly from Equation 1 Prove it by a combinatorial argument Hint : partition in two parts the set of subsets of size k ; those containing a given element and those not.

5 / 23 Binomial Coefficients and Subsets Enumeration

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PASCAL’S TRIANGLE(2)

1 1 1 1 2 + 1 1 3 + 3 + 1 1 4 + 6 + 4 + 1 1 5 + 10 + 10 + 5 + 1 1 6 + 15 + 20 + 15 + 6 + 1 1 7 + 21 + 35 + 35 + 21 + 7 + 1 1 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 1 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 1 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 1 11 + 55 + 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1 1 12 + 66 + 220 + 495 + 792 + 924 + 792 + 495 + 220 + 66 + 12 + 1 1 13 + 78 + 286 + 715 + 1287 + 1716 + 1716 + 1287 + 715 + 286 + 78 + 13 + 1 1 14 + 91 + 364 + 1001 + 2002 + 3003 + 3432 + 3003 + 2002 + 1001 + 364 + 91 + 14 + 1 1 15 + 105 + 455 + 1365 + 3003 + 5005 + 6435 + 6435 + 5005 + 3003 + 1365 + 455 + 105 + 15 + 1 1 16 + 120 + 560 + 1820 + 4368 + 8008 + 11440 + 12870 + 11440 + 8008 + 4368 + 1820 + 560 + 120 + 16 + 1

Thanks to Tikz/Gaborit 6 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

7 / 23 Binomial Coefficients and Subsets Enumeration

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THE BINOMIAL THEOREM

For all integer n and a formal parameter X (1 + X)n =

n

• k=0

n k

• X k (Newton 1666)

(6)

8 / 23 Binomial Coefficients and Subsets Enumeration

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THE BINOMIAL THEOREM

For all integer n and a formal parameter X (1 + X)n =

n

• k=0

n k

• X k (Newton 1666)

(6) Prove it by a combinatorial argument

8 / 23 Binomial Coefficients and Subsets Enumeration

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THE BINOMIAL THEOREM

For all integer n and a formal parameter X (1 + X)n =

n

• k=0

n k

• X k (Newton 1666)

(6) Prove it by a combinatorial argument Hint : write (1 + X)n = (1 + X)(1 + X) · · · (1 + X)

• n terms

in each term choose 1 or X, what is the coefficient of X k in the result (think "vector of n bits"). Exercises Use a combinatorial argument to prove :

n

• k=0

n k

• = 2n

Use the binomial theorem to prove (give also a combinatorial argument)

n

• k=0 k odd

n k

• =

n

• k=0 k even

n k

• = 2n−1

8 / 23 Binomial Coefficients and Subsets Enumeration

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SUMMATIONS AND DECOMPOSITIONS

The Vandermonde Convolution For all integers m, n, k

k

• j=0

m j

• n

k − j

• =

m + n k

• (7)

Prove it by a combinatorial argument Hint : choose k elements in two sets one of size m and the other n. Exercise Prove that

n

• k=0

n k 2 = 2n n

• (8)

Hint : Specify Equation 7

9 / 23 Binomial Coefficients and Subsets Enumeration

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SUMMATIONS AND DECOMPOSITIONS (2)

Upper summation For all integers p n

n

• k=p

k p

• =

n + 1 p + 1

• (9)

Exercises Establish the so classical result

n

• k=1

k 1

• Compute

n

• k=2

k 2

• and deduce the value of

n

• k=1

k2

10 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

11 / 23 Binomial Coefficients and Subsets Enumeration

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THE MAIN RULES IN COMBINATORICS (I)

Bijection Rule Let A and B be two finite sets if there exists a bijection between A and B then |A| = |B| . Summation Rule Let A and B be two disjoint finite sets then |A ∪ B| = |A| + |B| . Moreover if {A1, · · · An} is a partition of A (for all i = j, Ai ∩ Aj = ∅ and n

i=0 Ai = A)

|A| =

n

• i=0

|Ai| .

12 / 23 Binomial Coefficients and Subsets Enumeration

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THE MAIN RULES IN COMBINATORICS (II)

Product rule Let A and B be two finite sets then |A × B| = |A| . |B| . Inclusion/Exclusion principle Let A1, A2, · · · An be sets |A1 ∪ · · · ∪ An| =

n

• k=1

(−1)k

• S⊂{1,··· ,n}, |S|=k
• i∈S

Ai

• .

Exercises Illustrate these rules by the previous examples, giving the sets on which the rule apply.

13 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

14 / 23 Binomial Coefficients and Subsets Enumeration

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DERANGEMENT

Definition A derangement of a set S is a bijection on S without fixed point. Number of derangements of n elements dn (notation !n).

15 / 23 Binomial Coefficients and Subsets Enumeration

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DERANGEMENT

Definition A derangement of a set S is a bijection on S without fixed point. Number of derangements of n elements dn (notation !n). Inclusion/Exclusion principle !n = n! − n 1

• (n − 1)! +

n 2

• (n − 2)! − · · · + (−1)nn

n

• (n − n)!,

= n!

n

• i=0

(−1)i i!

n→∞

∼ n! 1 e . Recurrence relation Show by a combinatorial argument that dn = (n − 1)(dn−1 + dn−2) = ndn−1 + (−1)n.

15 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

16 / 23 Binomial Coefficients and Subsets Enumeration

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FIBONACCI NUMBERS

Recurrence Equation

• F0 = F1 = 1

Fn = Fn−1 + Fn−2 for all n 2

17 / 23 Binomial Coefficients and Subsets Enumeration

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FIBONACCI NUMBERS

Recurrence Equation

• F0 = F1 = 1

Fn = Fn−1 + Fn−2 for all n 2 Interpretation What kind of situation could be represented by Fibonacci’s Numbers ?

17 / 23 Binomial Coefficients and Subsets Enumeration

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FIBONACCI NUMBERS

Recurrence Equation

• F0 = F1 = 1

Fn = Fn−1 + Fn−2 for all n 2 Interpretation What kind of situation could be represented by Fibonacci’s Numbers ? Hint : Consider words in {0, 1}n Use a combinatorial argument to prove Fn = Fn−2 + Fn−3 + · · · + F1 + F0

17 / 23 Binomial Coefficients and Subsets Enumeration

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FIBONACCI NUMBERS

Recurrence Equation

• F0 = F1 = 1

Fn = Fn−1 + Fn−2 for all n 2 Interpretation What kind of situation could be represented by Fibonacci’s Numbers ? Hint : Consider words in {0, 1}n Use a combinatorial argument to prove Fn = Fn−2 + Fn−3 + · · · + F1 + F0 Hint : Consider the last "1" Imagine other combinatorial equalities

17 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

18 / 23 Binomial Coefficients and Subsets Enumeration

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PIGEONS AND HOLES

Principle If you have more pigeons than pigeonholes Then some hole must have at least two pigeons Generalization If there are n pigeons and t holes, then there will be at least one hole with at least n t

• pigeons

History Johann Peter Gustav Lejeune Dirichlet (1805-1859) Principle of socks and drawers

http://www-history.mcs.st-and.ac.uk/Biographies/Dirichlet.html 19 / 23 Binomial Coefficients and Subsets Enumeration

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IRRATIONAL APPROXIMATION

Friends Let α be a non-rational number and N a positive integer, then there is a rational p

q

satisfying 1 q N and

• α − p

q

• 1

qN Hint : divide [0, 1[ in N intervals, and decimal part of 0, α, 2α, · · · , Nα Sums and others

1

Choose 10 numbers between 1 and 100 then there exist two disjoint subsets with the same sum.

2

For an integer N, there is a multiple of N which is written with only figures 0 and 1

Geometry

1

In a convex polyhedra there are two faces with the same number of edges

2

Put 5 points inside a equilateral triangle with sides 1. At least two of them are at a distance less than 1

3

For 5 point chosen on a square lattice, there are two point such that the middle is also on the lattice

20 / 23 Binomial Coefficients and Subsets Enumeration

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GRAPHS

Friends Six people Every two are either friends or strangers Then there must be a set of 3 mutual friends or 3 mutual strangers Guess the number Player 1 : pick a number 1 to 1 Million Player 2 Can ask Yes/No questions How many questions do I need to be guaranteed to correctly identify the number ? Sorting

21 / 23 Binomial Coefficients and Subsets Enumeration

SUBSETS ALGEBRA RULES DERANGEMENT FIBONACCI PIGEONS’ HOLES REFERENCES

BINOMIAL COEFFICIENTS AND COMBINATORICS

1

THE PROBLEM : Subset Enumeration

2

ALGEBRAIC APPROACH

3

COMBINATORIAL RULES

4

DERANGEMENT

5

FIBONACCI’S NUMBERS

6

PIGEONS’ HOLES

7

BIBLIOGRAPHY

22 / 23 Binomial Coefficients and Subsets Enumeration

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REFERENCES I

Martin Aigner and Günter M. Ziegler. Proofs from THE BOOK. Springer, 8 2014. Robert A. Beeler. How to Count : An Introduction to Combinatorics and Its Applications.

Springer, 2015.

Alan Camina and Barry Lewis. An Introduction to Enumeration (Springer

Undergraduate Mathematics Series). Springer, 2011.

Philippe Flajolet and Robert Sedgewick. Analytic Combinatorics. Cambridge

University Press, 2009.

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. Concrete

Mathematics : A Foundation for Computer Science (2nd Edition). Addison-Wesley Professional, 1994.

Richard P. Stanley. Enumerative Combinatorics, Volume 2. Cambridge University Press,

1999.

Richard P. Stanley. Enumerative Combinatorics : Volume 1 (Cambridge Studies in

Advanced Mathematics). Cambridge University Press, 2011.

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