1 Lecture 4 (part 2) Combinatorics Reading: Epp Chp 6.

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Lecture 4 (part 2)

CombinatoricsReading: Epp Chp 6

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Outline

1. Basic Rulesa) Linear Series Rule

b) Multiplication Rule

c) Addition and Difference Rule

d) Inclusion-Exclusion Rule

2. Common Counting Scenariosa) Permutations (Ordered Selections)

b) Combinations (Unordered Selections)

c) Permutations with repetitions

d) Combinations with repetitions

3. Algebra of Combinations

4. Binomial Theorem

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2.1 Permutations

Definition: – An r-permutation of a set of n elements is

an ordered selection of r elements taken from the set of n elements (1 r n).

Notation:– The number of r-permutations of a set of n

elements is denoted as P(n,r).

Note: permutations are ordered selections: selecting the object in the first try/position, is considered different from selecting that object in the second or subsequent tries/positions.

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2.1 Permutations Theorem:

– If n and r are integers such that 1 r n, thenP(n,r) = n(n-1)(n-2)…(n-r+1)

or equivalentlyP(n,r) = n! / (n-r)!

Proof:– Uses the multiplication rule.– Formal proof is by induction. Informal proof is as follows:

1 r n{x1, x2 , x3 , … , xn}

Number of choices in step i is n-(i-1), regardless of the choices made in previous steps => Can use Multiplication Rule: n(n-1)(n-2)…(n-r+1)

Step 1: n choices

Step 2: n–1 choices

Step r : n–(r–1) choices

n!(n-r)!

n(n-1)(n-2)…(n-r+1)(n-r)(n-r-1)(n-r-1) …3.2.1(n-r)(n-r-1)(n-r-1) …3.2.1

=

Position 1 Position 2 Position r, , ,

Result: Ordered Tuple

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2.1 Permutations Theorem:

– If n and r are integers such that 1 r n, thenP(n,r) = n(n-1)(n-2)…(n-r+1)

or equivalentlyP(n,r) = n! / (n-r)!

Corollary: P(n,n) = n! / (n-n)! = n!

Theorem: CIRCULAR PERMUTATIONS– The number of ways of permuting r objects from a set of n objects in a

circle (in which two arrangements are the same when one is a rotation of the other) is P(n,r) / r

– When n = r, we have P(n,n) / n = (n-1)!

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2.1 Permutations

Example 1: (a) How many different ways can three of the letters of the word BYTES be chosen and written in a row? (b) How many different ways can this be done if the first letter must be ‘B’

Answer:

(a) The problem is a 3-permutation from a set of 5 objects.

– P(5,3) = 5!/2! = 60

(b) 2 step process (Multiplication Rule)– Step 1: Assign B to the first position: 1 way

– Step 2: Assign 4 letters to the remaining 2 positions: P(4,2) ways, regardless of step 1.

– Answer = 1 x P(4,2) = 12.

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2.1 PermutationsExample 2:

(a) In how many ways can the letters in the word COMPUTER be arranged in a row?(b) In how many ways can the letters in the word COMPUTER be arranged in CO must appear together in order?

Answer:

(a) P(8,8) = 8! (permutate all eight objects)

(b) P(7,7) = 7! (Treat CO as a unit)

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2.1 Permutations

Example 3: (a) How many ways can a group of 6 people be seated around a table?(b) How many ways can a group of 6 people be seated around a table when two of them cannot sit next to each other?

Answer:

(a) P(6,6)/6 = 5! (circular permutation)

(b) By difference rule:

Number of ways a group of 6 can sit around a table

– Number of ways in which the 2 enemies sit next to each other

= 5! – (2xP(5,5)/5) (treat the 2 enemies as 1 person)

= 5! – (2x4!)

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2.2 Combinations

Definition: – An r-combination of a set of n elements is

a subset of r elements taken from the set of n elements (r n).

Notation:– The number of r-permutations of a set of n

elements is denoted as C(n,r), also asn

r

Note: A combination is an unordered selection: you are selecting a ‘set’, and ordering is not important in sets.

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2.2 Combinations Theorem:

– If n and r are integers such that r n, thenC(n,r) = P(n,r) / r!

or equivalentlyC(n,r) = n! / ( r!(n-r)! )

Informal Proof:

{x1, x2 , x3 , … , xn}

Position 1 Position 2 Position r, , ,

, , ,

Step 1: Ordered selection: P(n,r)

Step 2: Get rid of duplicates since ordering is not important: P(n,r) / r!

Result: Ordered Tuple

Result: Set

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2.2 CombinationsExample 1:

You are to select five members from a group of twelve to form a team.(a) How many distinct five-person teams can be selected?(b) If two of them insist on working together as a pair, such that any team must either contain both of neither. How many five person teams can be formed?

Answer:

(a) C(12,5)

Set of twelve people

Set of five people

(b) Use Addition Rule:All 5-person teams satisfying the ‘2-friends’

constraint.

Those teams which

involve the 2 ‘friends’

Those teams which do not involve the 2 ‘friends’

Step 1: select the two ‘friends’: 1 way

Step 2: select the remaining 3 people: C(10,3)

C(10,3) C(10,5)+

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2.2 Combinations

Example 2: You are to select five members from a group of twelve to form a team. If two of them insist on working apart,how many five person teams can be formed?

Answer: 1st version, using difference rule.

All 5-person teams

Those teams with

the two enemies together

Those teams with the two enemies apart

Step 1: select the two enemies: 1 way

Step 2: select the remaining 3 people: C(10,3)

C(10,3)

C(12,5)

Ans: C(12,5) – C(10,3)

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2.2 Combinations

Example 2: You are to select five members from a group of twelve to form a team. If two of them insist on working apart,how many five person teams can be formed?

Answer: 2nd version, using addition rule.

Those teams which

contain A but not B

All 5-person teams that do not contain the two enemies, say A and B

Those teams which

contain B but not A

Those teams

which do not contain

A nor B

Step 1: select the A: 1 way

Step 2: select the remaining people except B: C(10,4)

C(10,4) C(10,4) C(10,5)+ +

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2.2 CombinationsExample 3: A group of twelve consists of five men and seven women.

(a) How many five-person teams can be chosen that consists of three men and two women?(b) How many five-person teams contain at least one man?(c) How many five-person teams contain at most one man?

(a) Answer:Step 1: choose the men: C(5,3) waysStep 2: choose the women: C(7,2) ways (regardless of the choices made in step 1)

Multiplication rule: C(5,3) x C(7,2)

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(b) Answer:Number of 5-person teams that contain at least one man= Number of 5-person teams– Number of 5-person teams that do not contain any men (all women).

(DIFFERENCE RULE)

= C(12,5) – C(7,5)

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(b) Answer:Number of 5-person teams that contain at most one man= Number of 5-person teams that contain no men+ Number of 5-person teams that contain 1 man

(BY ADDITION RULE)

= (C(5,0) x C(7,5)) Step 1: Choose 0 men from 5 menStep 2: Choose 5 women from 7 women

+ (C(5,1) x C(7,4)) Step 1: Choose 1 man from 5 menStep 2: choose 4 women from 7 women

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2.2 CombinationsExample 4: 10 people are to sit around two tables in a hawker (dealer) centre. The first table has 6 chairs,

the second table has 4 chairs.(a) How many ways can this be done?(b) How many ways can this be done if two of them needs to sit together?

(a) Answer (version#1):Step 1: Choose 6 people to sit on the first table

C(10,6)Step 2: Arrange the 6 people on the first table

5! (Circular Permutation)Step 3: Arrange the remaining 4 on the second table.

3! (Circular Permutation)(BY MULTIPLICATION RULE)

= C(10,6) x 5! x 3!= 10!/24

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2.2 CombinationsExample 4: 10 people are to sit around two tables in a hawker centre. The first table has 6 chairs, the

second table has 4 chairs.(a) How many ways can this be done?(b) How many ways can this be done if two of them needs to sit together?

(a) Answer (version#2):Step 1: Permute 6 people from 10 circularly around the first table

P(10,6)/6Step 2: Arrange the remaining 4 on the second table.

3! (Circular Permutation)(BY MULTIPLICATION RULE)

= P(10,6)/6 x 3!= 10!/24

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(b) Answer (version#1):BY ADDITION RULENumber of sitting arrangements where the two sit on the 1st table

+ Number of sitting arrangements where the two sit on the 2nd table

BY MULTIPLICATION RULEStep 1: Put the two on the first table: 1 wayStep 2: Select 4 more to join them: C(8,4)Step 3: Permute them around the table : 5!Step 4: Permute the remaining 4 circularly on 2nd table: 3!

BY MULTIPLICATION RULEStep 1: Put the two on the second table: 1 wayStep 2: Select 2 more to join them: C(8,2)Step 3: Permute them around the table: 3!Step 4: Permute remaining 6 circularly on 1st table: 5!

= (C(8,4)x5!x3!) + (C(8,2)x3!x 5!)

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(b) Answer (version#2):BY ADDITION RULENumber of sitting arrangements where the two sit on the 1st table

+ Number of sitting arrangements where the two sit on the 2nd table

BY MULTIPLICATION RULEStep 1: Put the two on the first table: 1 wayStep 2: Permute 4 from 8 circularly on the 2nd table: P(8,4)/4Step 3: Permute the 6 circularly on 1st table: 5!

BY MULTIPLICATION RULEStep 1: Put the two on the second table: 1 wayStep 2: Permute 6 from 8 circularly on the 1st table: P(8,6)/6Step 3: Permute remaining 4 circularly on 2nd table: 3!

= (P(8,4)/4 x 5!) + (P(8,6)/6 x 3!)

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2.3 Permutations with repetitions Theorem (Permutations from a multi-set):

Given a collection of n objects of which:– n1 are of type 1 and are indistinguishable from each other;– n2 are of type 2 and are indistinguishable from each other;

…

– nk are of type k and are indistinguishable from each other;

and that n1 + n2 + … + nk = n, then the number of distinct permutations of the n objects is

C(n, n1) C(n – n1 , n2) C(n – n1 – n2 , n3) … C(n – n1 – n2 – … – nk-1 , nk)or equivalently

n! / (n1! n2! … nk! )

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2.3 Permutations with repetitions

Example 1:

How many 8-bit strings have exactly three 1’s?

Answer (version#1 – details): Example:

– 10001101 is an 8-bit string, but has four 1’s– 10010001 is an 8-bit string and has exactly three 1’s.

TECHNIQUE: Sometimes, you have to know (a) the set you are choosing from; and (b) the destination ‘type’ that you are choosing to.– Point of view#1: Choose from the set {0,1} and place into the bit

positions 1,2,3,4,5,6,7,8.– Point of view#2: Choose from the bit positions 1,2,3,4,5,6,7,8, and

‘place’ them into the set {0,1} (i.e. assign them to either a bit 0 or 1) Adopting different points of view of the same problem may help. Use point of view#2:

– Step 1: Choose 3 bit positions from 8 to be assigned a bit 1: C(8,3)– Step 2: Choose the remaining 5 bit positions for bit 0: C(5,5) = 1

Answer: C(8,3) (By multiplication rule)

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Example 1:

How many 8-bit strings have exactly three 1’s?

Answer (version#2 – direct application of formula): What are the objects that you are permuting

– Point of view#1: 1’s and 0’s with duplicates?– Point of view#2: the bit positions

Again, it is still using point of view#2.

Permute 8 objects (the bit positions) – 3 of which are indistinguishable from each other– 5 of which are indistinguishable from each other

Answer: 8!/(3! x 5!)

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Example 2:

How many ways are there of ordering the letters of the word MISSISSIPPI, which are distinguishable from each other?

Answer: What are the objects that you are permuting

– Point of view#1: The Letters M,I,S,P? (with duplicates?)– Point of view#2: The 11 positions where the letters can be placed

Again, point of view#2. Permute 11 objects (the letter positions)

– 4 of which are indistinguishable from each other (for the letter ‘S’)– 4 of which (a different type) are indistinguishable from each other (for

the letter ‘I’)– 2 of which (a different type) are indistinguishable from each other (for

the letter ‘P’)– 1 position for letter ‘M’

Answer (use formula): 11!/(4! x 4! x 2! x 1!) Answer (from scratch): C(11,4) x C(7,4) x C(3,2) x C(1,1)

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2.4 Combinations with repetitions Definition (r-combinations from a multi-set):

Given a set X of n objects, an r-combination with repetition allowed (or r-combination with a multi-set of size r) is an unordered selection of elements taken from X with repetition allowed.

Theorem: The number of r-combination with repetition allowed drawn from a set of n elements is C(r+n–1 , r).

Example: 3-combinations from a set {a,b,c,d}– [a,a,a]; [a,a,b]; [a,a,c]; [a,a,d]– [a,b,b]; [a,b,c]; [a,b,d];– [a,c,c]; [a,c,d]; [a,d,d];– [b,b,b]; [b,b,c]; [b,b,d];– [b,c,c]; [b,c,d]; [b,d,d];– [c,c,c]; [c,c,d]; [c,d,d];– [d,d,d]

Total Number = 20 = C(3+4-1 , 3)

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Example: 3-combinations from a set {a,b,c,d}

Category a Category b Category c Category d

[a,a,a] X X X

[a,a,b] X X X

[a,b,d] X X X

[a,c,d] X X X

[b,c,c] X X X

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[a,a,a] X X X

[a,a,b] X X X

[a,b,d] X X X[a,c,d] X X X

[b,c,c] X X X

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[a,a,a]

[a,a,b]

[a,b,d]

[a,c,d]

[b,c,c]

Selection

X X X | | |

X X | X | |

X | X | | X

X | | X | X

| X | X X |

Represented by:Problem generalized:

• r-combination = putting r crosses

• From a set of n elements = putting n-1 ‘|’s in between crosses.

Reduces to the same problem of assigning 2-bits (X and |) to r+n-1 positions. (Permuting r+n-1 positions from a multi-set of {X,|}

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2.4 Combinations with repetitionsExample 1: A person giving a party wants to set out 15 assorted cans of soft drinks for his guests.

He shops at a store that sells 5 different types of soft drinks(a) How many different selections of cans of 15 soft drinks can he make?(b) If root beer is one of the types of soft drinks, how many different selections include at least 6 cans of root beer?

Answer:(a) 15-combination (15 drinks) from a multi-set of 5 categories

– r = 15 drinks = 15 crosses– n = 5 categories: need 4 ‘|’ to separate the crosses– C(15+5-1 , 15) = C(19, 15) = 3876

(b) Step 1: take out 6 cans of root beer: 1 wayStep 2: select the remaining 9 cans: 9-combination (9 remaining drinks) from a multi-set of 5 categories: C(9+5-1 , 9) = C(13,9) = 715

Final Answer = 1 x 715 = 715 (Multiplication Rule)

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2.4 Combinations with repetitions

Example 2: How many solutions are there to the equation

x1 + x2 + x3 + x4 = 10

(a) if x1, x2, x3, x4 are non-negative integers?

(b) if x1, x2, x3, x4 are positive integers?

Answer:(a) 10-combination from a multi-set of 4 categories

– r = 10 units = 10 crosses; to be distributed into…– n = 4 categories: the four variables: need 3 ‘|’ to separate the crosses– C(10+4-1 , 10) = C(13, 10) = 286

(b) Step 1: Assign 1 unit to each of the four variables: 1 way.Step 2: Assign the remaining 6 units to the four variables:

6-combination from a multi-set of 4 categories= C(6+4-1 , 6) = C(9,6) = 84

Final Answer = 1 x 84 = 84 (Multiplication Rule)

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2.4 Combinations with repetitions

Example 3: How many triplets of the form (i,j,k) are there, when

(a) each i , j , k {1,…,n}

(b) each i , j , k {1,…,n} and 1 i j k n

Answer:(a) Permutation: n3

(b) 3-combinations from a multi-set of n categories:(Observation skills needed to relate problem to known scenario)

For example if 1 i j k 5, then we have 3-combinations from a multi-set of 5 categories: 3 X’s and 4 ‘|’s

(1,1,2) being represented as XX|X|||(1,2,4) being represented as X|X||X|(2,3,5) being represented as |X|X||X

Therefore in general, answer is C(n+3-1 , 3)= (n+2)!/ (3! x (n-1)!)= (n+2)(n+1)n / 6

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Outline

1. Basic Rulesa) Linear Series Rule

b) Multiplication Rule

c) Addition and Difference Rule

d) Inclusion-Exclusion Rule

2. Common Counting Scenariosa) Permutations (Ordered Selections)

b) Combinations (Unordered Selections)

c) Permutations with repetitions

d) Combinations with repetitions


4. Binomial Theorem

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Common properties(1) C(n,r) = C(n, n-r)(2) Pascal’s Formula: C(n+1, r) = C(n, r-1) + C(n,r)

Proof of (1) Proven Algebraically:

C(n,r) = n! / ( r! (n-r)! ) = n! / ( (n-r)! (n-(n- r))! ) = C(n, n-r) Proven using combinatorial reasoning (p331 of text)

Let A be a set with n elements.

Let the subsets of A of size r be B1, B2,…,Bk.

Each Bk can be paired up with a subset of A of size n–r: namely A – Bk.

All subsets of size n All subset of size n–r

B1 A – B1

B2 A – B2

… …

Bk A – Bk

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Common properties(1) C(n,r) = C(n, n-r)(2) Pascal’s Formula: C(n+1, r) = C(n, r-1) + C(n,r)

Proof of (2): refer to p334 of recommended text. Implications: Pascal’s Triangle

00

10

11

21

22

21

31

32

30

33

41

42

40

43

44

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

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4. Binomial Theorem

Binomial Theorem:

(a+b)n = an-kbk

k=0

nnk

Algebraic proof is by induction on n. (p341 of text) Combinatorial Proof:

(a+b)n = (a+b) (a+b) (a+b) … (a+b) (n copies) #1 #2 #3 #n

b a a … a

a b a … a

• What is the coefficient of an-kbk ?• How many ways can we create the an-kbk term?• Well, to create a term, we have to select either a or b from

each (a+b) group. • Problem reduces to a n-permutation with repetitions

(previous section 2.3), of 2 types of objects: selecting n-k copies of a and k copies of b: n!/(k!(n-k)!) = C(n,k)

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4. Binomial Theorem

Binomial Theorem provides rapid expansion of polynomials.– Example 1: Expand (a+b)n.

C(5,0)a5b0 + C(5,1)a4b1 + C(5,2)a3b2 + C(5,3)a2b3 + C(5,4)a1b4 + C(5,5)a0b5

= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

It also allows you to access a particular term without the need to expand the entire polynomial– Example 2: What is the coefficient of a9b6

in the expansion of (a+b)15?• Ans: C(15,6)

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4. Binomial Theorem

Corollary of Binomial Theorem:2n = (1+1)n

= 1n-k1k (by binomial thm)k=0

nnk

n0

n1

n2

nn= + + + +

k=0

nnk=

Q: If |S| = n, what is |P(S)|?

A: By addition rule:

number of subsets of size 0 + number of subsets of size 1+ number of subsets of size 2 + … + number of subsets of size

n

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End of Lecture

1 Lecture 4 (part 2) Combinatorics Reading: Epp Chp 6.

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