Slides Chapter2 Counting

7/25/2019 Slides Chapter2 Counting

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GEK1505: Chapter 2 Counting

Chapter 2 Counting 1 / 127
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Overview

1 Geometric series

2 Arithmetic series

3 SetsUnion, intersection & complementPrinciple of inclusion & exclusion

Examples

4 Counting principlesAddition principleMultiplication rinciple

5 Arrangements & Combinations

6 Number of routes on rectangular grid

7 Pigeonhole Principle

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Geometric series

Definition (Geometric series)

A geometric series is a sum of the form

a+ar+ar2 +ar3 + +arn.



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Geometric series

Theorem

The value of a geometric series is given by the formula

a+ar+ar2 +ar3 + +arn = a(rn+1 1)r 1 .

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Proof. Let

Sn = a+ar+ar2 +ar3 + +arn. (1)

Then,

rSn = ar+ar2 +ar3 +ar4 +

+arn+1. (2)

Subtracting (1) from (2), we have

rSn Sn = arn+1 a

Sn = a(rn+1

1)

r 1 .



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Geometric series

Example (Tower of Hanoi)

There are three vertical poles A, B, C. Five circular slabs of different radiiand each with a hole in the centre rest on top of each other as shown inthe figure below. What is the minimum number of steps needed totransfer all the slabs to pole Bif at each step no slab is allowed to rest on

top of a smaller slab in the same pole?

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Geometric series

Solution.Suppose there are only 2 slabs.

We need minimum number of 3 steps:

Move the top (smallest) slab to pole C;Move the bottom (larger) slap to pole B;

Move the smaller slab from pole C to pole B.

Let an be the minimum number of steps required to move a tower ofn

slabs. Thena2= 3.



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Geometric series

Suppose there are n slabs.

Let us label them as 1, 2, etc from the top (smallest) slab to the bottom

(largest) slab.When can we move the n-th slab to pole B?

This happens exactly when the first n 1 forms a tower in pole C. The(minimum) number steps to achieve this is an1.



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Geometric series

It takes an1 steps to move the first n 1 slabs to pole C;It takes 1 step to move the n-th slab from pole A to pole B;

It takes anothera

n

1 steps to move then

1 slabs from poleC

topole A.

Hence,

an = 1 + 2an1.



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Geometric series

In our case, n= 5, so

a5 = 1 + 2a4

= 1 + 2(1 + 2a3)

= 1 + 2 + 22a3

= 1 + 2 + 22

(1 + 2a2)= 1 + 2 + 22 + 23a2

= 1 + 2 + 22 + 23(1 + 2a1)

= 1 + 2 + 22 + 23 + 24

= 1(25

1)2 1

= 31.



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Arithmetic series



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Arithmetic series

The numbers given by T1, T2, etc are called the triangular numbers.

T1 = 1T2 = 1 + 2 = 3

T3 = 1 + 2 + 3 = 6

T4 = 1 + 2 + 3 + 4 = 10

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Arithmetic series

Theorem

The triangular numberTn is given by

Tn = 1 + 2 + 3 + +n =12n(n+ 1).

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Arithmetic series

Proof.

We rearrange the terms:

Tn = 1 + 2 + +n 1 +nT

n = n+n

1 +

+ 2 + 1

Adding them:

2Tn = n(n+ 1)

Tn = n(n+ 1)

2

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Arithmetic series

More generally, an arithmetic series is a sum of the form

a+ (a+d) + (a+ 2d) + + (a+nd).

The number a is called the first term and the number d is called thecommon difference of the arithmetic series.

The triangular number Tn is the arithmetic series with

a=d= 1.

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Arithmetic series

Theorem

The sum of an arithmetic series is given by

a+ (a+d) + (a+ 2d) + + (a+nd) =(n+ 1)(2a+dn)2

.


A i h i i


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Arithmetic series

Proof.a+ (a+d) + (a+ 2d) + + (a+nd)

= (a+ +a) +d(1 + 2 + +n)= (n+ 1)a+dTn

= (n+ 1)a+dn(n+ 1)

2

= (n+ 1)(2a+dn)

2


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Arithmetic series

Proof.

Alternatively, we rearrange the terms:

An = (a) + (a+d) + + (a+nd)An = (a+nd) + (a+ (n

1)d) +

+ a

Adding them:

2An = (n+ 1)(2a+nd)

An = (n+ 1)(2a+nd)

2


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Arithmetic series

Example

1 + 3 + 5 + + (2n 1) =n2

.

1 + 4 + 7 + + (3n 2) = 12n(3n 1).


A ith ti i
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Arithmetic series

Example

Draw n lines in the plane.This will form a number of mutuallynon-overlapping region in the plane. What is the maximum number ofsuch regions that can be formed?

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Solution.

n = 0 1 2 3 4 . . . nRegions = 1 2 4 7 11 . . . ?

Plus 1 2 3 4 . . . n

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Here is an explanation.

Say when you have 2 lines and want to add a third line. This new thirdline intersects the existing lines and is divided into 3 segments. Eachsegment slices an existing region into 2 and thus we get 3 addition regions.

Now if you n 1 lines and you want to add the nth line. This new line isdivided into n segments and thus we have n additional regions.

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In general, for n lines, we get

1 + 1 + 2 + 3 + 4 + +n = 1 + (1 + 2 + +n)= 1 +

1

2n(n+ 1)

= 1

2(n2 +n+ 2).


Sets
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Sets

Similar objects are often gathered together for easy reference.

Such a collection is called aset.

Examples:

set of all NUS students

set of positive integers


Sets
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Sets

The items in a set are often referred to as elements ormembers of the

set.Sets are often denoted by uppercase letters like A, B, C etc.

Notation for sets

We exhibit members of a set within parentheses, e.g.

S={a, e, i, o, u} or

S={x : x is a vowel of the English alphabet}.We use the notation x S to mean x is a member ofS or xbelongsto S.

xS means x is not a member ofS.


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Sets

Definition (Empty set)

The empty set is the set containing no members. This is denoted by.That is ={}.

Think ofas an empty box.


Sets


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Sets

Definition (Union)

The union of sets A and Bis the set whose elements are precisely thosebelong to A orB. Symbolically, we denote the union by A

B:

A B={x :x A or x B}.

We can perform the union operation repeatedly to obtain A

B

C, etc.


Sets
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Sets

Example

IfA={a, b, c} and B={b, c, d, e}, thenA B={a, b, c, d, e}.


Sets
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Sets

Definition (Intersection)

The intersection of sets A and Bis the set whose elements are preciselythose belong to A and B. Symbolically, we denote the intersection by

A B:A B={x :x A and x B}.

We can perform the intersection operation repeatedly to obtain A

B

C

etc.


Sets
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Sets

Definition

We say that the sets A and B aredisjoint if

A B=.


Sets
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Sets

We can useVenn diagrams to visualise the union and intersection of sets:


Sets
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Theorem (Properties of union & intersection)A B=BA, A B=BA (Commutative law)(A B) C =A (B C) (Associative law)

(A B) C =A (B C) (Associative law)A (BC) = (A B) (A C) (Distributive law)A (BC) = (A B) (A C) (Distributive law)

A A= A, A A= A (Idempotent laws)A =A, A =


Sets
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Example

Give examples of sets A, B, C such that

(A B) C=A (BC).

Solution. Take

A={1, 2, 3}, B={2, 5}, C ={3, 5}.

Then

(A B) C={3, 5}.A (B C) ={1, 2, 3, 5}.


Sets
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Definition (Subset)

If every element ofA is also an element ofB, then we say that A is a

subset ofB.Symbolically, we write

AB.


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Some immediate consequences of the definition of a subset:

For any set A, AA, A.

For sets A, B,AA B, A B A.

IfAB, thenA B=B, A B=A.


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We are often interested in subsets of a fixed reference set called theuniversal set.

Definition (Complement)

Suppose S is the given universal set, and AS. Then the complementofA, denoted by A

c

, is the set consisting of all the elements ofS whichare not in A. That is

Ac ={x S :xA}.

By definition,A Ac =S, A Ac =.


Sets


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Theorem (de Morgans laws)

The following properties hold for setsA, B.

(A B)c =Ac Bc.

(A B)c =Ac Bc.


Principle of Inclusion & Exclusion
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Theorem (PIE for 2 sets)

For arbitrary finite setsA, B, C, we have

|A B|=|A| + |B| |A B|.

The principle is obvious by examining the Venn diagram.


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Example

At Dunman High School there are

28 students in algebra class,30 students in biology class, and

8 students in both classes. How many students are in either algebraor biology class?


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Solution. LetA= set of students in algebra class

B= set of students in biology class

Then|A B|=|A| + |B| |A B|= 28 + 30 8 = 50.


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Theorem (PIE for 3 sets)

For arbitrary finite setsA, B, C, we have

|AB C|=|A| + |B| + |C| |AB| |A C| |B C| + |AB C|.


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Proof.

|A B C| = |(A B) C|= |A B| + |C| |(A B) C| (3)


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Applying PIE for 2 sets, we have

|A B| = |A| + |B| |A B| (4)

|(A

B

) C

| = |(A

C

) (B

C)|

= |A C| + |B C| |(A C) (B C)|

= |A C| + |B C| |(A B C| (5)

The result follows by substituting (4, 5) into (3).


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Example

At Dunman High School there are

55 students in either algebra, biology, or chemistry class

28 students in algebra class

30 students in biology class

24 students in chemistry class

8 students in both algebra and biology

16 students in both biology and chemistry

5 students in both algebra and chemistryHow many students attend all three classes?


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Solution. Let A= set of students in algebra class

B= set of students in biology class

C= set of students in chemistry class

Then

|A B C|= 55

|A

|= 28,

|B

|= 30,

|C

|= 24

|A B|= 8, |B C|= 16, |A C|= 5.


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By PIE,

|AB C|=|A| + |B| + |C| |AB| |A C| |B C| + |AB C|.

|A BC| = |A B C| |A| |B| |C|+|A B| + |A C| + |BC|

= 55 28 30 24 + 8 + 16 + 5= 2.

There are 2 students who attend all three classes.


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Example (PIE for 4 sets)

For arbitrary sets A, B, C, D, we have

|A B C D|=|A| + |B| + |C| + |D||A B| |A C| |A D| |BC| |BD| |C D|+|A BC| + |A BD| + |A C D| + |B C D||A BC D|.


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Solution.

By PIE for 2 sets,

|(A

B

C)

D

| =

|A

B

C

|+

|D

| |(A

B

C)

D

|By the distributive law,

|(A

B

C)

D

| =

|(A

D)

(B

D)

(C

D)

|


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Apply the PIE for 3 sets to calculate

|A

B

C|, |

(AD)

(B

D)

(C

D)

|and rearrange the terms.


PIE: Examples
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ExampleA pavement was tiled by 80 parallel slabs of tiles. A cat walked along thepavement and at every fifth tile other than the first) i.e. at tiles numbered5, 10, 15, ..., it relieved itself.

A while later, another cat came along and did the same thing but only atevery seventh tile (other than the first), i.e. at tiles numbered 7, 14, 21, ...

Later a boy strolled along the pavement, each step covering theodd-numbered tiles.

How many of the tiles covered by his steps will not have been soiled byany of the two cats?


PIE: Examples


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Solution. Consider the following sets:

A={1, 2, 3, . . . , 80}, B={5, 10, 15, . . . , 80}

C={

7, 14, 21, . . . , 77}, D=

{1, 3, 5, . . . , 79

}.

BC= Set of tiles soiled by any of the two cats(BC)c = Set of ttiles not soiled by any of the two cats

Goal: Find|D (BC)c|.


PIE: Examples
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Use the fact that for any set X,

D= (D X) (D Xc).

Take X =B C. Then

|D (B C)c|=|D| |D (B C)|.

Next, determine|D| and|D (BC)|.


PIE: Examples
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Clearly,|D|= 40.

|D (B C)| = |(D B) (D C)| (by distributive law)= |D B| + |D C| |D BC| (by PIE)

= 8 + 6 1= 13.

since

|D B| = |{5, 15, 25, 35, 45, 55, 65, 75}|= 8|D C| = |{7, 21, 35, 49, 63, 77}|= 6

|D B C| = |{35}|= 1


PIE: Examples
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Finally,

|D

(B

C)c

| =

|D

| |D

(B

C)

|= 40 13= 27.


Addition principle
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Addition principle of counting

If a choice from set A1 can be made in n1 ways, and a choice from set A2can be made in n2 ways, then the number of choices from A1 orA2 is

n1+n2.

Necessary condition: Set A1 and A2 are disjoint.


Addition principle
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Addition principle of counting extension

If a choice from set Aican be made in niways for i= 1, . . . ,m, then thenumber of choices from A1 Am is

n1+ +nm.

Necessary condition: The sets A1, . . ., Am are pairwise/mutuallydisjoint, i.e. Ai

Aj =

for all i

=j.


Addition principle
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Example

Given three setsA={a,m, r}B={b, d, i, l, u}C ={c, e, n, t}

How many ways are there to choose one letter from among the set A, BorC?

Solution. The three sets are pairwise/mutually disjoint. By the addition

principle, the number of ways is

3 + 5 + 4 = 12.


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Example

Here are two sets of positive integers:

A={2, 3, 5, 7, 11, 13}

B={2, 4, 6, 8, 10, 12}How many ways are there to choose one integer from A B?

The two sets are NOT disjoint. What modification can we make to the

addition principle to accommodate this case?


Multiplication principle
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Multiplication principle of counting

If a task involves a sequence of two steps, where the first step can becompleted in n1 ways and the second step can be completed in n2 ways,then there are

n1 n2ways to complete the task.

Necessary condition: The ways each step can be completed areindependentof each other.


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Multiplication principle of counting extension

If a task involves a sequence ofm steps, where the i-th step can becompleted in niways, then there are

n1 nmways to complete the task.

Necessary condition: The ways each step can be completed are

independentof each other.


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Example

If a word is defined as a juxtaposition of letters, how many

(i) 2-letter words

(ii) 3-letter words

(iii) k-letter words

can be formed from the English alphabet?


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Solution. Suppose we wish to form a k-letter word, where k1. Thereare

26 ways to choose the 1st letter;

26 ways to choose the 2nd letter;...

26 ways to choose the k-th letter.

Each steps above are independent of each other. By the multiplicationprinciple, the total number of ways is

26 26 k times

= 26k.


Arrangements in a row

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Example

In how many ways can n different objects be arranged in a row?

Solution. We choose objects to place in a row one by one from left toright:

There are n ways for the first position;

There are n 1 ways for the 2nd position;...

There are 2 ways for the n 1-th position;There is 1 way for the n-th position.

By multiplication principle, the number of ways is

n! =n (n 1) 2 1.


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Example

In how many ways can you form a row of 3 different objects from 8different objects?

Solution. There are

8 choices for the 1st position;

7 choices for the 2nd position;

6 choices for the 3rd position.

The number of ways is8 7 6 = 336.


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Theorem

Let nPkbe the number of ways of arranging in a rowkdifferent objectstaken from n different objects. Then

nPk= n!

(n k)! .


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ExampleIn how many ways can you arrange 3 red, 4 green and 5 blue beads in arow?

Solution. If all the beads were different, then the number of ways is

(3 + 4 + 5)! = 12!.

Imagine we have such an arrangement.

Crucial observation: Permuting the beads of the same colour will yieldthe same arrangement!




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There are 3! ways to permute the red beads;

There are 4! ways to permute the green beads;

There are 5! ways to permute the blue beads.

By multiplication principle, the number of ways to arrange 3 red, 4 greenand 5 blue beads in a row is

12!

3!4!5!.


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Theorem

The number of ways of arranging in a rown1 identical objects of Type1,n2 identical objects of Type2, ..., nkidentical objects of Typek is equal to

(n1+n2+ +nk)!n1!n2! nk! .


Circular arrangements

Example
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Example

In how many ways can you arrange n different objects in a circle?

Solution.

A circular arrangement is considered the same after rotating it.


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There is no first position.

Any object can be used as a point of reference.

Important: Once a reference object is fixed, the order in which theremaining objects are arranged with respect to this reference object (eitherin clockwise or anti-clockwise direction) is now important.


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Suppose the objects are labelled as A1, A2, . . ., An.

Fix A1 as the reference object.

Arrange the rest in clockwise direction starting from A1.


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There are

(n 1) ways to choose the object next (in clockwise direction) to A1.(n 2) ways to choose the next object (in clockwise direction)....

2 ways to choose the second last object.

1 way to choose the last object.

Therefore, the total number of ways is

(n 1)! = (n 1) (n 2) 2 1.


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Theorem (Circular arrangements)

The number of ways of arrangingn different objects in a circle is

(n 1)!.


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ExampleFind the number of ways, without removing any empty seat, that 9 guestscan be seated at a round table with exactly 10 seats?

Solution. The empty seat could be used as the reference object. Thenthe number of ways is just

(10 1)! = 9!.


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Example

Find the number of ways, without removing any empty seat, that 8 guestscan be seated at a round table with exactly 10 seats?

Solution. Just imagine, in addition to the existing guests, we have twodifferent ghosts who will occupy the empty seats. Then the the number ofseatings is

(10

1)! = 9!.



Now, let us identify the two ghosts. If there were not identical, there arel i l i h h i i f h
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exactly twocircular arrangements with the same positions for theseghosts! But we want to identify these ghosts as identical since theycorrespond to empty seats.


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Therefore, the total number of ways is

9!

2 =

9

8

7!

2 = 36 7!.


Combinations
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Many counting problems involve counting the number of subsets of a finiteset satisfying certain constraints.

How many gift hampers can be designed to contain 10 out of 15given different items?

How many 5-person committees can be formed from a group of 8persons?

Given n different objects, how many ways can we choose kof them?

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Theorem

Let nCkdenote the number of ways of choosingkobjects from a set ofndifferent objects. Then

nCk

= n!

k!(n k)!.


Combinations
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Proof. For each such a collection ofkobjects, we can arrange them in arow, and there are k! ways to that.

By multiplication principle,

n

Ckk! =

n

PkThus,

nCk=nPk

k! =

n!

k!(n

k)!

.


Combinations
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Another notation for nCk is n

k

=

n!

k!(n k)!

We pronounce this number as

n choose k.


Combinations
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The numbers nk also called binomial coefficients since they appear inthe binomial expansion of (1 +x)n:

(1 +x)2 = 1 + 2x+x2

(1 +x)3 = 1 + 3x+ 3x2 +x3

(1 +x)4 = 1 + 4x+ 6x2 + 6x3 +x4

...

(1 +x)

n

= n

0+ n

1x+ n

2x2 + + n

nxn


Combinations
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Theorem (Simple properties of binomial coefficients)n0

=

nn

= 1

nk

=

nnk

n0+ n1+ n2+ + n

n= 2n

Proof. The first two are obvious from the definition.

The last one follows from the binomial expansion of (1 + x)n by setting

x= 1.


Combinations
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Example

How many different 5-member subcommittees can be formed from a9-member specified committee if the President and Vice-President do notserve on the same subcommittee?

Solution. There are 3 types of subcommittees:

Type 1: containing President but not Vice President

Type 2: containing Vice President but not President

Type 3: not containing President and Vice President


There are
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74 different subcommittees of Type 1.74

different subcommittees of Type 2.7

5

different subcommittees of Type 3.

By addition principle, the total number is7

4

+

7

4

+

7

5

= 91.


Combinations
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Example

An LCD display consists of 10 rows and 10 columns of squares which maybe lighted up.

How many different patterns of lighted display can be obtained if 30squares are lighted up?

How many of these have at least one lighted square touching the border ofthe display?


Combinations
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There are 100 squares altogether.

If exactly 30 squares are lighted up, there are

10030

different patterns.


Combinations
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Suppose none of the 30 squares touch the border. The number of these is8 8

30

=

64

30

.

Those have at least one lighted square touching the border is100

30

64

30

.


Rectangular grid
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ExampleIf you are allowed to go east or north, how many ways can you go from Ato B in the following figure:


Rectangular grid

A possible route is given in the following figure in red colour:
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Convert this path into a string of integers consisting of 0s and 1s asfollows:

Horizontal step corresponds to 0

Vertical steps corresponds to 1.


Rectangular grid

The 01-string corresponds to the path is
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0 1 0 0 1 0 1 0 0.

The length of this 01-string is always 9, which is the sum of thedimensions of the grid.

The dimension of the grid here is 3

6. For this grid, the corresponding01-string always have

Six 0s

Three 1s

Conversely, any 01-string of length 9 will give a path from A to Bconsisting of east or north bound move at each step.


Rectangular grid
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The required number of different routes is equal to the number of different01 string of length 9, with six 0 and three 1.

Hence, the number required is given by

9!

6!3!.


Rectangular grid

We label each intersection point on a rectangular grid using coordinate(i,j) such that i increases from left to right, and j increases from bottom
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to top.

The bottom left point is labelled by (0, 0).


Rectangular grid
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Theorem

On a rectangular grid, the number of routes from(0, 0) to(i,j) movingeasterly or northerly without back-tracking is

(i+j)!

i!j! =

i+j

i

=

i+j

j

.


Rectangular grid
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More generally,

Theorem

On a rectangular grid, the number of routes from(i,j) to(k, l), where

i k and j l, moving easterly or northerly without back-tracking is((k i) + (lj))!

(k i)!(lj)! =k+l ij

k i

=

k+l ij

lj.


Rectangular grid
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Example

How many possible bus routes (moving easterly or northerly without

back-tracking) are there from (0, 0) to (8, 8) that pass through (3, 3) or(5, 5)?


Rectangular grid
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Solution. Let

A= set of routes from (0, 0) to (8, 8) and passing through (3, 3).

B= set of routes from (0, 0) to (8, 8) and passing through (5, 5)

Aim: Compute

|A

B

|.

Cha te 2 Co ti g 97 / 127

Rectangular grid
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|A| =

6

3

10

5

= 5040

|B| = 105 6

3= 5040

|A B| =

6

3

4

2

6

3

= 2400.

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Rectangular grid
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By PIE,

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

= 5040 + 5040 2400= 7680.

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Example

How many possible bus routes (moving easterly or northerly without

back-tracking) are there from (0, 0) to (7, 8) thatdoes notpass throughany of the points (3, 3) or (5, 5)?

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Rectangular grid

Solution.
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A= set of routes from (0, 0) to (7, 8) and passing through (3, 3).

B= set of routes from (0, 0) to (7, 8) and passing through (5, 5).

X= set of routes from (0, 0) to (7, 8).

Aim: Compute|X| |A B|.

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|X| =

15

7

= 6435

6 9
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|A| = 63 9

4= 2520|B| =

10

5

5

2

= 2520

|A B| =6

3

4

2

5

2

= 1200

|X| |A B| = |X| |A| |B| + |A B|= 2595.


Pigeonhole Principle
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Example

If 3 objects are distributed among 2 pigeon-holes, then there is at least one

pigeon hole that will contain at least two of the distributed objects.


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Theorem (Pigeon-hole Principle (PHP))Supposem objects are distributed amongn pigeon-holes. Ifm>n, thenthere is at leastone pigeon-hole with at leasttwoof the distributedobjects.

Proof. If every pigeon-hole contains at most one of the distributedobjects, then the total number of the distributed objects is at most n,contradicting the fact that there are more than n objects being

distributed.


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When solving problems using PHP, we need to identify

what are the pigeon-holes

what are the distributed objects



Example
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Jacky takes lunch at the same cafeteria five days a week (Monday toFriday).

There are 16 different food-stalls.

Within how many weeks will it be certain that Jacky will take his lunch atleast twice from the same food-stall?

Distributed objects Pigeon-holes

days food-stalls


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By PHP, a minimum of 17 days guarantees that Jack will order from thesame food-stall on at least two days.

Thus within 4 weeks, it is certain that he will take his lunch at least twicefrom the same food-stall.


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ExampleIf you are at a reception attended by more than 366 guests, you can besure that there will be at least two persons with the same birthday.


guests birthdays



Example

A fitness club classifies its members according to the following
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characteristics: gender, age, height and weight, for which there are 2, 8, 5and 5 classes respectively.

If the club has more than 400 members, then there will be at least 2members with exactly the same characteristics.


members characteristics

Total number of characteristics:

2 8 5 5 = 400.


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ExampleA dating agency requires each of its clients to answer a questionnaireconsisting of 10 questions with yes/no options only.

There are 525 male and 500 female clients.

Are the following statements true?

(i) There will be two questionnaires forms with identical answers.

(ii) There will be one male and one female whose questionnaires answersare identical.



Solution.
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(i) Yes


clients (525 + 500 = 1025) answers (210 = 1024)

(ii) No.

It is possible that the identical answers come from both male or both

female clients.


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Theorem (Extended PHP)

Ifm objects are distributed amongn pigeon-holes andm>n, then therewill beone pigeon-hole which containsat least

m/n

objects.

Here,m/n denotes the smallest integerm/n.


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Proof.Let k=m/n. Then k 1< m/nk ornk n


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ExampleThe members of a youth club are grouped into age groups according to x:

16x


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diagonals in the ratio of 2 to 1 (see figure below), and with an area of 680sq km.

Partition the island into N2 equal rhombic regions by drawing lines parallelto the sides of the rhombus respectively in such a way that each side of

the rhombus is divided into Nequal segments.

Assume that there are 3.5 million residents in Singapore. Let N= 10.

(i) There is at least one rhombic region with population density of at

least 35000/6.8 residents per sq km.

(ii) In a gathering of more than 100 residents, there will be at least twopersons who stay within 5.2km of each other.


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Let x(in km) be the length of the shorter diagonal of a rhombic region.Then

N2x2 = 680.



(i)
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residents (3, 500, 000) rhombic regions (102 = 100)

By extended PHP, there is at least one region with at least

3500000

100 = 35000

residents in it.


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The area of each region is

x2 = 680/100 = 6.8 sq km

There is one region with population density35000

6.8

residents per sq km.


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(ii) By PHP, in a gathering of more than 100 residents, there will at leasttwo persons who reside in the same region.

The longest diagonal of the region is

2x= 26.85.2 km.

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Example

In any group ofn people there are at least two persons having the samenumber friends. (It is assumed that if a person x is a friend ofy then y isalso a friend ofx.)


Proof.
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The number of friends of a person x is an integer kwith 0k n 1.If there is a person ywhose number of friends is n 1, then everyone is afriend ofy, that is, no one has 0 friend.

This means that 0 and n

1 can not be simultaneously the numbers offriends of some people in the group. The pigeonhole principle tells us thatthere are at least two people having the same number of friends.


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ExampleAmong three given integers, there must be two of them whose sum is even.


different integers given (3) parity of integer even or odd (2)

By PHP, there will be two integers with the same parity, and so their sum

must be even.



Example

Classes are conducted from 8 : 00 am to 5 : 00 pm from Monday to Friday
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in one-hour time slots.

You have 25 hours of classes each week.

Which of the following statements is true?

(a) There is one day during which you have at least 5 hours of classes.

(b) There is one day during which you have at least 6 hours of classes.

(c) There are at least 2 days in which you have classes with a commontime slot.

(d) There are at least 3 days in which you have classes with a commontime slot.


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Solution.


classes (25) days (5)

(a) True.(b) False.



Di ib d bj Pi h l
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classes (25) time slots per day (9)

By extended PHP, there is one time slot with at least

25/9 + 1 = 3 classes.

(c) True.(d) True.



E l
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ExampleProve that if seven distinct numbers are selected from{1, 2, . . . , 11}, thensome two of these numbers sum to 12.

Solution. Let

A1={1, 11}, B={2, 10}, C={3, 9}, D={4, 8},E ={5, 7}, F ={6}.


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the seven distinct number selected the sets A, B, C, D, E, F

By PHP, two of the seven selected numbers must come from the same set.This set cannot be F; so it must be one of the sets A, B, C, D, E. In allcases, these two numbers sum to 12.

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Documents

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