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Reminder. 3 rd and last test: Date : Saturday, 19 th Oct Time : 1430hrs Topics: Lectures 7-10c Contributes to 20% of CA. Prepare for test as if you’re preparing for the finals. Lecture 11 (part 2). Combinatorics Reading: Epp Chp 6. Outline. Basic Rules Linear Series Rule

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Reminder

Reminder

  • 3rd and last test:

    • Date : Saturday, 19th Oct

    • Time : 1430hrs

    • Topics: Lectures 7-10c

    • Contributes to 20% of CA.

    • Prepare for test as if you’re preparing for the finals.


Lecture 11 part 2

Lecture 11 (part 2)

Combinatorics

Reading: Epp Chp 6


Outline

Outline

  • Basic Rules

    • Linear Series Rule

    • Multiplication Rule

    • Addition and Difference Rule

    • Inclusion-Exclusion Rule

  • Common Counting Scenarios

    • Permutations (Ordered Selections)

    • Combinations (Unordered Selections)

    • Permutations with repetitions

    • Combinations with repetitions

  • Algebra of Combinations

  • Binomial Theorem


2 1 permutations

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.


2 1 permutations1

Result: Ordered Tuple

{x1, x2 , x3 , …, xn}

Step 1: n choices

Step 2: n–1 choices

Step r : n–(r–1) choices

,

,

,

Position 1

Position 2

Position r

n!

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

=

(n-r)!

(n-r)(n-r-1)(n-r-1) …3.2.1

2.1 Permutations

  • Theorem:

    • If n and r are integers such that 1£ r £n, then

      P(n,r) = n(n-1)(n-2)…(n-r+1)

      or equivalently

      P(n,r) = n! /(n-r)!

  • Proof:

    • Uses the multiplication rule.

    • Formal proof is by induction. Informal proof is as follows:

1£ r £ n

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)


2 1 permutations2

2.1 Permutations

  • Theorem:

    • If n and r are integers such that 1£ r £n, then

      P(n,r) = n(n-1)(n-2)…(n-r+1)

      or equivalently

      P(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)!


2 1 permutations3

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.


2 1 permutations4

2.1 Permutations

Example 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)


2 1 permutations5

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!)


2 2 combinations

n

r

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 as

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


2 2 combinations1

Result: Ordered Tuple

{x1, x2 , x3 , …, xn}

,

,

,

Position 1

Position 2

Position r

Result: Set

,

,

,

2.2 Combinations

  • Theorem:

    • If n and r are integers such that r £n, then

      C(n,r) = P(n,r) / r!

      or equivalently

      C(n,r) = n! /( r!(n-r)! )

  • Informal Proof:

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

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


2 2 combinations2

Set of twelve people

Set of five people

C(10,3)

2.2 Combinations

Example 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?

(b) Use Addition Rule:

Answer:

(a) C(12,5)

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

+

C(10,5)

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


2 2 combinations3

All 5-person teams

C(12,5)

C(10,3)

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.

Those teams with the two enemies together

Those teams with the two enemies apart

Step 1: select the two enemies: 1 way

Ans: C(12,5) – C(10,3)

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


2 2 combinations4

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

C(10,4)

C(10,5)

C(10,4)

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

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)


2 2 combinations5

2.2 Combinations

Example 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(5,2) ways

Step 2: choose the women: C(7,2) ways (regardless of the choices made in step 1)

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


2 2 combinations6

2.2 Combinations

Example 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?

(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)


2 2 combinations7

2.2 Combinations

Example 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?

(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


2 2 combinations8

2.2 Combinations

Example 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#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


2 2 combinations9

2.2 Combinations

Example 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)/6

Step 2: Arrange the remaining 4 on the second table.

3! (Circular Permutation)

(BY MULTIPLICATION RULE)

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

= 10!/24


2 2 combinations10

2.2 Combinations

Example 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?

(b)Answer (version#1):

BY ADDITION RULE

Number 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 RULE

Step 1: Put the two on the first table: 1 way

Step 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 RULE

Step 1: Put the two on the second table: 1 way

Step 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!)


2 2 combinations11

2.2 Combinations

Example 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?

(b)Answer (version#2):

BY ADDITION RULE

Number 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 RULE

Step 1: Put the two on the first table: 1 way

Step 2: Permute 4 from 8 circularly on the 2nd table: P(8,4)/4

Step 3: Permute the 6 circularly on 1st table: 5!

BY MULTIPLICATION RULE

Step 1: Put the two on the second table: 1 way

Step 2: Permute 6 from 8 circularly on the 1st table: P(8,6)/6

Step 3: Permute remaining 4 circularly on 2nd table: 3!

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


2 3 permutations with repetitions

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!)


2 3 permutations with repetitions1

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)


2 3 permutations with repetitions2

2.3 Permutations with repetitions

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!)


2 3 permutations with repetitions3

2.3 Permutations with repetitions

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)


2 4 combinations with repetitions

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)


2 4 combinations with repetitions1

Category a

Category b

Category c

Category d

X X

X

X

X

X

X

X

X

X

X X

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]

X X X

[a,a,b]

[a,b,d]

[a,c,d]

[b,c,c]


2 4 combinations with repetitions2

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]

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


2 4 combinations with repetitions3

X X X | | |

X X | X | |

X | X | | X

X | | X | X

| X | X X |

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}

  • 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,|}

Selection

Represented by:

[a,a,a]

[a,a,b]

[a,b,d]

[a,c,d]

[b,c,c]


2 4 combinations with repetitions4

2.4 Combinations with repetitions

Example 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 way

    Step 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)


2 4 combinations with repetitions5

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, x4are non-negative integers?

(b) if x1, x2, x3, x4are 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)


2 4 combinations with repetitions6

2.4 Combinations with repetitions

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

(a) each i , j , kÎ {1,…,n}

(b) each i , j , kÎ {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


Outline1

Outline

  • Basic Rules

    • Linear Series Rule

    • Multiplication Rule

    • Addition and Difference Rule

    • Inclusion-Exclusion Rule

  • Common Counting Scenarios

    • Permutations (Ordered Selections)

    • Combinations (Unordered Selections)

    • Permutations with repetitions

    • Combinations with repetitions

  • Algebra of Combinations

  • Binomial Theorem


3 algebra of combinations

3. Algebra of Combinations

Common properties

  • C(n,r) = C(n, n-r)

  • 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

    … …

    BkA – Bk


3 algebra of combinations1

4

1

0

1

2

2

2

3

3

3

4

4

4

4

3

0

1

1

1

0

2

2

2

1

0

1

0

3

4

3

1

1

1

2

1

1

3

3

1

1

4

6

4

1

3. Algebra of Combinations

Common properties

  • C(n,r) = C(n, n-r)

  • 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

1


4 binomial theorem

n

k=0

n

k

4. Binomial Theorem

  • Binomial Theorem:

    (a+b)n = an-kbk

  • 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 ofan-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)


4 binomial theorem1

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)


4 binomial theorem2

n

n

k=0

k=0

= 1n-k1k(by binomial thm)

=

n

n

n

n

n

n

0

k

n

k

1

2

¼

=

+

+

+

+

4. Binomial Theorem

  • Corollary of Binomial Theorem:

    2n= (1+1)n

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


Reminder

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