01 Combinatoriek I

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Combinatorics

Wiskundige Methoden / Mathematical Methods FEB21010(X)

dr. Twan [email protected] - H11-10

Econometric Institute

May 1, 2013

Twan Dollevoet (Econometric Institute) Combinatorics May 1, 2013 1 / 27


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What is combinatorics?

Combinatorics is about counting!

Counting all possible license plates

Counting the different hands in a card game

Counting the number of passwords for a website

Counting the possibilities to arrange books on a book shelf

Counting the ways to distribute bonuses over employees

. . .



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Example: PasswordsA password for a website should satisfy the following conditions.

It should consist of 12 characters.

It should consist of letters and digits only.

Question: How many different passwords are possible.

1 {a, b , . . . , z , A, B , . . . , Z , 0, 1, . . . , 9} ↔ {1, 2, . . . , 62}.

2 Choosing a password is equivalent to selecting 12 times a numberbetween 1 and 62.

3 Answer:

6212 = 3, 226, 266, 762, 397, 899, 821, 056 ≈ 3.3 · 1021

4 Two important observations:◮ Repetition is allowed: AAABBBCCCAAA is a valid password.◮ Order is important: ABCDEFGHIJKL = LKJIHGFEDCBA.



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Example: Counting the number of hands in a poker gameTexas Hold’em

The game is played with a standard deck containing 52 cards.

Every player gets two cards from the deck.

Then, the game starts.

Examples of hands: {Ace♥, Ace♠}, {4♦, Jack♣}

Question: How many different hands are possible?

Two important observations:

Repetition is not allowed: You cannot have {Ace♠, Ace♠}.

Order is unimportant: {4♦, Jack♣} = {Jack♣, 4♦}.

Further questions:

What is the probability of getting two Aces?

What is the probability of getting two Hearts

. . .Twan Dollevoet (Econometric Institute) Combinatorics May 1, 2013 4 / 27


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Four fundamental counting problems

Four basic counting problems can be classified according to

whether repetition is allowed;

whether the order is important.

Repetition is Repetition is

k from n not allowed allowedOrder is k -permutation k -permutation with

important from n repetition from n

Order is not k -combination k -combination withimportant from n repetition from n



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Repetition is Repetition is

k from n not allowed allowedOrder is k -permutation k -permutation with

important from n repetition from npasswords


hands in a card game



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Examples for remaining fundamental counting problems

1 Committee to organize a party, consisting of

◮ a chairman, and◮ a treasurer, and◮ a secretary.

Characteristics: No repetition, order is important.

Question: If there are 10 candidates, how many committees can be formed?

2 A grandmother wants to distribute 20 chocolates over 10grandchildren.

Characteristics: With repetition, order is unimportant.

Question: In how many ways can grandmother distribute the chocolates over her grandchildren?



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Basic counting problem: Select k times from a set of n elements.

Four variants can be classified according towhether repetition is allowed;


Repetition is Repetition isk from n not allowed allowed

Order is k -permutation k -permutation withimportant from n repetition from n

forming a committee passwords


hands in a card game distributing chocolate



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What is counting?

Basic rule 1.0

Counting is determining the number of elements in a set.

We denote the number of elements in a set A by |A|.

The number of elements in a set A equals the number of ways tochoose an element x ∈ A.

One-to-one rule 1.3Let A and B be finite sets. If and only if there is a one-to-onecorrespondence between A and B , then the number of elements in A andB is equal: |A| = |B |.

A one-to-one correspondence is also called a bijection.Examples:

◮ {A, B , . . . , Z } ↔ {1, 2, . . . , 26}◮ Deck of cards ↔ {1, 2, 3, 4} × {1, 2, . . . , 13} ↔ {1, 2, . . . , 52}.

We denote the set {1, 2, . . . , n} by N n.



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whether repetition is allowed;whether the order is important.



forming a committee passwords???





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Our first result

Theorem 1.4: k from n, with repetition, order is important

Selecting a k -permutation from n elements, with repetition, can be done in

nk

ways.

Examples:

The number of 12-character passwords over an alphabet of 62characters equals 6212 ≈ 3.3 × 1021.

The number of integers that a computer can represent by 8 bits (= 1byte) equals 28 = 256.

The number of three-letter ‘words’ on a license plate equals263 = 17576.



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Rule of sum and rule of product

Recall the following theorems from Precalculus.

Rule of sum 1.1

If A and B are finite, disjoint sets, then |A ∪ B | = |A| + |B |.

In general, if A ∪ B = ∅, then |A ∪ B | = |A| + |B | − |A ∩ B |.

Rule of product 1.2

If A and B are finite sets, then |A × B | = |A| · |B |.



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Example: Counting the possible license plates in Belgium

A license plate in Belgium consists of

a number between 1 and 7;three capital letters;

a number with three digits.

Examples: 1-ABC-123; 4-DKM-542; 3-ZZZ-999;

There are 100 combinations of three letters that cannot be used.Examples: AAP, BOM, DIK

Question: How many license plates are possible in Belgium?

Two important observations:Repetition is allowed: 1-TTT-111

Order is important: 1-ABC-123 = 1-CBA-213



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Example: Counting the possible license plates in BelgiumA license plate in Belgium consists of

a number between 1 and 7;

three capital letters;

a number with three digits.

Exactly 100 three-letter combinations are forbidden.

{License plates} ↔ X × Y × Z ,

where

X = {1, . . . , 7} = N 7,

Y = {three-letter, non-forbidden words},

Z = {three-digit number}.

|{License plates}|One-to-one

↑= |X ×Y ×Z |

Product rule↑= |X | · |Y | · |Z | = 7 ·(263−100) ·103.



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whether repetition is allowed;whether the order is important.



forming a committee passwords??? nk





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Example: Displaying products in a music store

Consider a music store with 50 DVD’s .

The store has room for displaying 10 DVD’s.(“Our recommendation” / “Our Top 10” / . . .)

Question: In how many ways can the shop display 10 DVD’s?

50 · 49 · . . . · 41 = 50 · 49 · . . . · 41 ·

40 · 39 · . . . · 1

40 · 39 · . . . · 1

= 50 · 49 · . . . · 41 · 40 · 39 · . . . · 1

40 · 39 · . . . · 1

= 50 · 49 · . . . · 1

40 · 39 · . . . · 1=

50!

40! =

50!

(50 − 10)!

Here n! = n · (n − 1) · . . . · 1.


O


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Our second result

Theorem 1.5: k from n, without repetition, order is important

Selecting a k -permutation from n elements, without repetition, can bedone in

P (n, k ) = n!

(n − k )!

ways.

Remarks:

Note that n! = n · (n − 1) · . . . · 1; 1! = 1; 0! = 1.

Arranging all n items can be done in

n!

(n − n)! =

n!

0! =

n!

1 = n!

ways.


F f d l i bl


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displaying DVD’s passwords

P (n, k ) = n!(n−k )! nk

Order is not k

-combination k

-combination withimportant from n repetition from nhands in a card game distributing chocolate


E i A i b k h lf


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Exercise: Arranging books on a shelf

A student possesses6 books on economics, and

6 books on mathematics.

The books are arranged in a sequence on a book shelf.

How many options are there, if

1 the books can be placed in any order?

2 the books on economics should be on the left?

3 all books on mathematics should be adjacent?

4 the subjects should alternate?



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I t S t diff


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Intermezzo: Set difference

Set difference A− B

Let A and B be sets. Then we define their difference as

A − B = {x ∈ A : x /∈ B }.

Alternative notation for the set difference is A \ B .

Examples: Consider the sets N 6 = {1, 2, 3, 4, 5, 6} and A = {1, 2, 3, 5}.Then

N n − A = {4, 6}.

Let B = {2, 5}. Then

A − B = {1, 3}.

Let C = {3, 4}. ThenA − C = {1, 2, 5}.


F f d t l ti g bl s


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P (n, k ) = n!(n−k )! nk


hands in a card game distributing chocolate???


Our third ‘result’


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Our third result

Important observation: Two similar problems

1

Selecting k

elements from a set of n

elements without repetition2 Selecting a subset of size k from a set with n elements.

Definition: k from n, without repetition, order is unimportant

We define

C (n, k ) =

n

k

as the number of k -combinations without repetition from n

We pronounce this as ‘n choose k ’, or ‘n over k ’

Tomorrow, we will derive an expression that allows us to compute thesenumbers!


Combinatorial theorems I


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Combinatorial theorems I

Theorem: Complementarity (Exercise 1.21)

For n ≥ 0, 0 ≤ k ≤ n, it holds that

n

k

=

n

n − k

.

Proof Consider the set N n.nk

is the number of subsets of N n of size k .

If A is a subset of size k , then N n − A is a subset of size n − k .

There is one-to-one correspondence between subsets of size k andsubsets of size n − k :

{subsets of size k } → {subsets of size n − k } : A → N n − A

By the one-to-one rule the number of subsets of size k equals thenumber of subsets of size n − k .


Combinatorial theorems II


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Combinatorial theorems II

Theorem: Pascal’s identity (Exercise 1.22)

For n ≥ 0, 1 ≤ k ≤ n, it holds thatn + 1

k

=

n

k

+

n

k − 1

.

Proof Consider the set N n+1.n+1k

is the number of subsets of N n+1 of size k .

Each subset A of size k satisfies either1 n + 1 /∈ A: A ⊆ N n: A is a subset of N n of size k 2 n + 1 ∈ A: A − {n + 1} ⊆ N n: |A − {n + 1}| is a subset of N n of size

k − 1.There is a one-to-one correspondence between

1 set of subsets of N n of size k or k − 1, and2 set of subsets of N n+1 of size k .




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P (n, k ) = n!(n−k )! nk


hands in a card game distributing chocolateC (n, k ) =

n

k


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01 Combinatoriek I