1 lecture 4 (part 1) combinatorics reading: epp chp 6

1

Lecture 4 (part 1)

CombinatoricsReading: Epp Chp 6

2

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) Counting ordered partitions / permutations of multi-sets

3. Algebra of Combinations

4. Binomial Theorem

3

1. Basic Rules

1.1 Linear Series Rule.

If m and n are integers such that m n, then there are n-m+1 integers from m to n inclusive.

Example 1: How many integers are there in the sequence 10, 11, 12, …, 19, 20?

Ans: 20 – 10 + 1 = 11

Example 2: How many integers are there in the sequence -8,-7,-6,…-1,0,1,2,…4, 5?

Ans: 5 – (–8) + 1 = 14

4

1. Basic Rules

1.1 Linear Series Rule.

If m and n are integers such that m n, then there are n-m+1 integers from m to n inclusive.

Example 3: How many integers are there in 0 to 1000 inclusive, that are divisible by 3.

The integers are 0,3,6,9,…,996,999 , which are in the form 3k, where k = 0,1,2,…,333.

Hence there are 333-0+1 = 334 integers from 0 to 1000 which are divisible by 3.

5

1. Basic Rules: Multiplication Rule

1.2 Multiplication Rule

If an operation consists of a sequence of steps/events E1, E2 … Ek and if each Ei can be performed in ni ways regardless of how the previous steps E1, … Ei-1 were performed (i.e. independent), then the entire operation can be performed in n1n2…nk ways.

Proof of multiplication rule is by induction

(Left as an exercise)

6


1.2 Multiplication Rule

If an operation consists of a sequence of steps/events E1, E2 … Ek and if each Ei can be performed in ni ways regardless of how the previous steps E1, … Ei-1 were performed (i.e. independent), then the entire operation can be performed in n1n2…nk ways.

Example 1:

At breakfast, you are given a choice of coffee or tea, and a choice of a scrambled, fried and boiled egg. How many different kinds of breakfast can you have?

Scrambled

Fried

Boiled

Scrambled

Fried

Boiled

Coffee

Tea

Ans:

Step 1: Choose coffee or tea => 2 ways

Step 2: Choose how you would like your egg to be done => 3 ways regardless of the choice made in step 1.

Total number of ways = 2 x 3 = 6 ways (Mul. Rule)

7


Example 2:

Let A and B be finite sets. If |A| = m and |B| = n, how many elements are there in (i) A B? (ii) P(A)?

(i) The task of selecting pairs for A B is reduced to the following steps:

Step 1: Choose an element x from A : m ways.

Step 2: Choose an element y from B : n ways regardless of choices made in step 1.

Total number of ways of performing the task = total number of elements in A B = m n (Multiplication Rule)

8


(ii) Let the elements of A be a1 , a2 , a3 ,…, am

The task of forming a subset of A is reduced to the following procedure:

Step 1: Either take or drop element a1 : 2 ways.

Step 2: Either take or drop element a2 : 2 ways regardless of choices made in step 1.

Step 3: Either take or drop element a3 : 2 ways regardless of choices made in steps 1 and 2.

…

Step m: Either take or drop element am : 2 ways regardless of choices made in steps 1, 2, 3, 4, …, m-1.

Total number of ways of performing the task (by multiplication rule) = 2 2 2 … 2

Example 2:

Let A and B be finite sets. If |A| = m and |B| = n, how many elements are there in (i) A B? (ii) P(A)?

m times of 2

= 2m

9


Example 3:

A car license plate has 3 letters of alphabet followed by 4 single digit numbers. How many different car licenses can be issued (a) if repetitions of alphabets/numbers in the license plate is allowed; (b) if repetitions are not allowed?

Ans (a):

The task of forming a license plate number consists of the following sub-tasks:Step 1: choose the 1st letter : 26 ways.

Step 2: choose the 2nd letter : 26 ways (independent of step 1).

Step 3: choose the 3rd letter : 26 ways (independent of steps 1-2).

Step 4: choose the 1st digit : 10 ways (independent of steps 1-3).

Step 5: choose the 2nd digit : 10 ways (independent of steps 1-4).

Step 6: choose the 3rd digit : 10 ways (independent of steps 1-5).

Step 7: choose the 4th digit : 10 ways (independent of steps 1-6).

Total number of performing the task = 263 104

10


Example 3:

A car license plate has 3 letters of alphabet followed by 4 single digit numbers. How many different car licenses can be issued (a) if repetitions of alphabets/numbers in the license plate is allowed; (b) if repetitions are not allowed?

Ans (b):

The task of forming a license plate number consists of the following sub-tasks:Step 1: choose the 1st letter : 26 ways.

Step 2: choose the 2nd letter : 25 ways (independent of step 1).

Step 3: choose the 3rd letter : 24 ways (independent of steps 1-2).

Step 4: choose the 1st digit : 10 ways (independent of steps 1-3).

Step 5: choose the 2nd digit : 9 ways (independent of steps 1-4).

Step 6: choose the 3rd digit : 8 ways (independent of steps 1-5).

Step 7: choose the 4th digit : 7 ways (independent of steps 1-6).

Total number of performing the task = 26 25 24 10 9 8 7

11


Example 4: (Limitations of the Multiplication Rule)

Two teams A and B are to play each other repeatedly until one wins two games in a row, or until a total of 3 games. How many ways can a tournament be played?

Incorrect answer:

Step 1: Either A wins or B wins : 2 ways.



Total number of ways = 2 2 2

dependent

dependent

12


Example 4: (Limitations of the Multiplication Rule)

Two teams A and B are to play each other repeatedly until one wins two games in a row, or until a total of 3 games. How many ways can a tournament be played?

The above example illustrates a situation when the multiplication rule cannot be used.

A wins B wins

A wins B wins

A wins B wins

A wins B wins

A wins B wins

This is because the subsequent events (games) are dependent on the outcome of the previous events.

13


Example 5: (Goal Re-ordering is a possible way to by-pass dependency of tasks)

Three officers – A president, a treasurer, and a secretary, are to be chosen from among 4 people: A, B, C, D. Suppose that A cannot be president, and either C or D must be secretary. How many ways can the officers be chosen?

Incorrect answer:

Selecting three officers can be broken down to the following tasks:

Step 1: Select the president : 3 ways.

Step 2: Select the treasurer : 3 ways.

Step 3: Select the secretary : 2 ways.

Total number of performing the task = 3 3 2 ways

dependent

dependent

14




C

D

A

Step 1: Select the president: 3 ways

Step 2: Select the treasurer: number of ways DEPENDENT on the outcome of step 1!!!

A

B

A

B

D

C

D

C

D

D

C

C

Step 3: Select the secretary: number of ways DEPENDENT on the outcome of step 2, which is in turn, DEPENDENT on step 1.

start

B

C

D

15




Answer:

Independence of events can sometimes be restored through the re-ordering of your tasks.

Step 1: Select the secretary : 2 ways (C or D).

Step 2: Select the president : 2 ways regardless of the choice taken in step 1 (person B with the remaining person from step 1)

Step 3: Select the treasurer : 2 ways. (person A with the person remaining from step 2)

Total number of performing the task = 2 2 2 ways

16




Step 1: Select the secretary: 2 ways

Step 2: Select the president: 2 ways

Step 3: Select the treasurer: 2 ways

B

D

B

C

A

D

A

BA

C

A

B

start

C

D

17


Summary: Always check that the number of ways of each

step is independent of the choices of the previous steps.

The multiplication rule did NOT say :

‘Each step is independent of the choices of previous step’.

It says :‘The number of ways of each step is independent of the choices of the previous step’

Get the subtle difference? Goal re-ordering may help.

18

1. Basic Rules: Addition/Difference Rule

1.3 Addition Rule:

If a finite set A = A1 A2 An where all the Ai’s are mutually disjoint, then |A| = |A1| + |A2| + + |An|

Difference Rule:

If A is a finite set and BA, then |A – B| = |A| – |B|

Addition Rule is proven by induction. (Proof are left as exercise).

Difference Rule is a corollary of the addition rule. (‘corollary’ meaning that ‘it is a result of’… meaning that you can prove/infer the difference rule from the addition rule.)

Because when BA, (A – B) and B form a partition of A(Meaning that (A – B) B = A and (A – B) B = )

Therefore by addition rule: |A – B| + |B| = |A|

And so: |A – B| = |A| – |B| A

B

19


Example 1: A computer access code word consists of one to three letters, chosen from the 26 alphabets, with repetitions allowed. How many code words are possible?

Answer: The set of all code words can be partitioned into subsets consisting

of those of length 1, length 2, and length 3.

Set of all code words of length 3




By addition rule: |Set of all code words of length 3|= | Set of all code words of length 1 | + | Set

of all code words of length 2 | + | Set of all code words of length 3 |

20


Example 1: A computer access code word consists of one to three letters, chosen from the 26 alphabets, with repetitions allowed. How many code words are possible?

Answer: The set of all code words can be partitioned into subsets consisting

of those of length 1, length 2, and length 3.





Number of code words of length 1 = 26 Number of code words of length 2 = 262 (Multiplication Rule) Number of code words of length 3 = 263 (Multiplication Rule) Total number of code words = 26 + 262 + 263 (Addition Rule)

21


Example 2: There are 15 different computer science books, 12 different math books, and 10 different chemistry books on the shelf. How many ways can we select 2 books, each from a different subject?

Answer:

Set of selections of 2 books

1 book from CS and 1 book Math

1 book from CS and 1 book from Chemistry

1 book from Math and 1 book from Chemistry

By addition rule: |Set of selections of 2 books| = | Set of selection of 1 book from CS and 1 book from Math | + | Set of selection of 1 book from CS and 1 book from Chem | + | Set of selection of 1 book from Math and 1 book from Chem |

22


Example 2: There are 15 different computer science boks, 12 different math books, and 10 different chemistry books on the shelf. How many ways can we select 2 books, each from a different subject?

Answer:

Set of selections of 2 books

1 book from CS and 1 book Math

1 book from CS and 1 book from Chemistry

1 book from Math and 1 book from Chemistry

15 12

(M.R.)

15 10

(M.R.)

12 10

(M.R.)

+ +

Addition rule is reasoning-by-cases applied to counting!

23


Example 3: A group of eight people are attending the movies together. If two of the eight people are enemies and do not want to sit next to each other, how many ways can this group sit in a row, such that the two enemies are separated?

Answer:

Different arrangements of 8 people in a row = 8!(by multiplicaton rule)

How?

Step 1: Sit the 1st person: 8 ways

Step 2: Sit the 2nd person: 7 ways regardless of outcome of step 1.

Step 3: Sit the 3rd person: 6 ways regardless of outcome of step 1-2.

…

Different arrangements

of 8 people in a row where the 2

enemies sit next to each

other

Different arrangements of 8 people in a row where the 2 enemies sit apart

24



Answer:

Different arrangements of 8 people in a row = 8!




other


= 27! (by multiplicaton rule)

How?

Step 1: Sit the 2 enemies together: 27 ways

Step 2-7: Sit the remaining 6 people: 654321

Step 1: Combine the 2 enemies as 1 person and take it as an arrangement of 7 people to 7 chairs, (7!).

Step 2: Different ways of arranging the 2 enemies to sit side-by-side: 2 ways.

OR… Another way of looking at it:

25



Answer:

Different arrangements of 8 people in a row = 8!




other


= 27! (by multiplicaton rule)

Answer = 8! – 27! (Difference Rule)

26


Example 4: How many integers are there in 1000 to 9999 that contain at least a digit 5.

Answer= Number of integers in 1000 to 9999 – Number of integers in 1000 to 9999 that do not contain a digit 5.

(Difference Rule)

= (9999 – 1000 + 1) – 8.93 (Linear Series Rule) (Multiplication Rule)

= 2439

27


Example 5: How many 3 digit numbers have at least one digit repeated?

Answer= Number of 3 digit numbers – Number of 3 digit numbers which have NO digit repeated

(Difference Rule)

(9 10 10) Multiplication Rule:Step 1: Choose hundreths digit (must

exclude leading ‘0’, therefore only 9 ways)Step 2: Choose tenths digitStep 3: Choose units digit

(9 9 8) Multiplication Rule:Step 1: Choose hundreths digit (must exlude leading

‘0’)Step 2: Choose tenths digit (10 ways, excluding

the digit in step 1. Therefore 9 ways)Step 3: Choose units digit

28

1. Basic Rules: Inclusion-Exclusion Rule

1.4 Inclusion-Exclusion Rule: (For 2 sets) Given any sets A and B,

|A B| = |A| + |B| – |A B| (For 3 sets) Given any sets A, B and C, |A

B C | = |A| + |B| + |C| – |A B| – |A C| – |B C| + |A B C|

(For n sets) Given any sets A1 … An, | A1 … An |

= | Ai |

– | Ai Aj |

+ | Ai Aj Ak |

– …

i 1..n

i,j 1..ni,j distinct

i,j,k 1..ni,j,k distinct

29


Moral of the story behind the inclusion-exclusion rule: – If you over-count, then subtract away

those you counted more than once.– If you under-count, then add back those

you have missed out.

This is applicable to any counting scenario. Inclusion-Exclusion Rule is the

generalized version of the addition rule. If the sets are disjoint, then the In-Ex rule reduces to the addition rule.

30


Example 1: How many integers from 1 through 1000 are multiples of 3 or 5? How many are neither multiples of 3 nor 5?

Answer:|{Integers of multiples of 3 or 5}|

= |{Integers of multiples of 3}|+ |{Integers of multiples of 5}| – |{Integers of multiples of 3 and 5}| (Inclusion-Exclusion Rule)

All integers from 1 to 1000

Multiples of 5

Multiples of 3

31



Answer:|{Integers of multiples of 3 or 5}|

= |{Integers of multiples of 3}|+ |{Integers of multiples of 5}| – |{Integers of multiples of 3 and 5}| (Inclusion-Exclusion Rule)

|{Integers of multiples of 3}| = |{3k | where k = 1,2,…333}| = 333|{Integers of multiples of 5}| = |{5k | where k = 1,2,…200}| = 200|{Integers of multiples of 3 and 5}| = |{15k | where k = 1,2,…66}| = 66

Ans = 333 + 200 – 66= 467

32



Question: How many are neither multiples of 3 nor 5?Answer: {Int from 1..1000 with are multiples of 3 or 5} {Int from 1..1000}

Ans = 1000-467 (By difference rule)= 533

467 1000

33

1. Basic Rules: Inclusion-Exclusion RuleExample 2: In a class of 50 students, 30 know Pascal, 18 know Fortran, 26 know COBOL, 9 know both Pascal and

Fortran, 16 know Pascal and COBOL, 8 know both Fortran and COBOL, 47 knows at least one of the three languages.(a) how many students know none of the three languages?(b) how many students know all three languages?(c) how many students know Pascal and Fortran but not COBOL?(d) how many students know Pascal, but neither Fortran nor COBOL?

P = set of students who know Pascal; C = set of students who know COBOLF = set of students who know Fortran; U = All 50 students.

(a) |U – (P C F)| = |U| – |(P C F)|(By difference rule, since (P C F) U)

|U| = 50 (Given)|(P C F)| = 47 (Given)

Therefore answer = 50-47 = 3.

34




(b) |(P C F)| = |P| + |C| + |F| – |PC| – |PF| – |CF| + |PCF|

(Inclusion-Exclusion Rule for 3 sets)47 = 30 + 26 + 18 – 9 – 16 – 8 + |PCF|

|PCF| = 6

35




(c) |((PF) –(PFC))| = |PF| – |PFC| (Difference Rule)= 9 – 6 = 3

P C

F

69

36




P C

F

36

10

30

?

(d)Similar to (c), those who know Pascal and COBOL, but not Fortran = 16 – 6 = 10.

Since |P| = 30By difference rule, those who knowPascal, but neither fortran nor COBOL= 30 – 19 = 11

37

Summary so far: Basic Counting Rules

Four basic Rules– Linear Series Rule– Multiplication Rule– Addition/Difference Rule– Inclusion-Exclusion Rule

All counting can be broken down to these four rules… they may be likened to the ‘atoms’ of counting.

IMPT: Just like the logical inference rules, you use and combine them together to create a proof, also here, you have to combine the use of the counting rules when necessary, to tackle the overall counting problem

We now use our ‘atoms’ to build common ‘procedural calls’ to tackle common counting scenarios:– Permutations– Combinations

38

End of Lecture

1 lecture 4 (part 1) combinatorics reading: epp chp 6

Documents