Math 103 04 Combinatorics

8/11/2019 Math 103 04 Combinatorics

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Math 103Statistics andProbability

Combinatorics

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Fundamentals

Experiment : Any process that generates a set of data.

Observations : The recorded information as a result of an experiment

Sample Space S : The set of all possible outcomes of an experiment.

Sample Point : A particular outcome in the sample space.

Event : A subset of a sample space.

Simple Event : An event with only one element (sample point).

Compound Event : The union of simple events.

Null Space : Subset of sample space containing no elements.

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Example

Experiment: Toss a coin thrice.

First Toss Second Toss Third Toss

Tree Diagram for listing all sample points of sample space S

H

T

T

T

T

T

T

T

H

H

H

H

H

H

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

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Example (cont)

SH2 = Event when exactly two heads appear= { HHT, HTH, THH }

S1H = Event when the first toss results in a head= { HHH, HHT, HTH, HTT }

SH = Event when all heads appear (a simple event)

= { HHH }

Venn Diagram

HHH

HHT

HTH

HTT

THH

TTT

THT

TTH

S

SH

S1H SH2


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SH2 = { HHT, HTH, THH }

S1H = { HHH, HHT, HTH, HTT }SH = { HHH }

Intersection : contains all sample points common to both events.A = SH2 S1H = { HHT, HTH }

Mutually Exclusive Events : cannot occur simultaneously.

B = SH2 SH = (SH2 and SH are mutually exclusive)

Union : contains all sample points from first OR second set.

C = SH2 S1H = {HHT, HTH, THH, HHH, HTT}

Complement : contains sample points not in given set

D = S1H = S S1H = { TTT , TTH, THT, THH }

Operations on Events

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Exercise

Experiment : Roll a pair of dice. (one red, one white)

What is the sample space?

Find A = event of getting a total of more than 10 dots.Find B = event of getting an odd sum

Find C = event of getting the same number of dots on each die

Find the intersection of A and B

Find the intersection of B and C

Find the union of B and CFind the complement of the intersection of A and B

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Combinatorics

Combinatorial Reasoning underlies analysis

in computer science (like algorithms),discrete operations, research problems and

finite probability.

Systematic analysis of different possibilities

Exploration of logical structure of a problem

Ingenuity

We are primarily concerned with

Counting Techniques.

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Counting PrinciplesThe Addition PrincipleIf there are n1 different objects in the first set, n2 objects inthe second, , nm objects in the k

th set,and if the different sets are disjoint (i.e. mutually exclusive orpairwise intersections are empty),

Then the number of ways to select an object from one ofthe m sets is n1+n2++nk.

The Multiplication Principle:If a first operation can be performed in n1 ways, and foreach of these a second operation can be performed in n2ways, , and for each way to perform the first (k-1)operations, the kth operation can be performed in nk ways.Then the number of ways to perform the k operations isn

1

n2

nk

.


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Addition Principle Example200 students take Calculus, 500 take Statistics and 150

take both, how many (distinct) students are in these twoclasses?

Solution 1: Use Inclusion-Exclusion formula from Discrete Math.

Solution 2 :

Addition rule cannot be applied directly because the sets are

not disjoint. So, categorize the data

200 150 = 50 are in Calculus but not in Statistics

500 150 = 350 are in Statistics but not in Calculus150 are in Calculus and Statistics.

These 3 events are disjoint the addition principle applies.

So the answer is

50 + 350 + 150 = 550

n (A B) = n (A) + n (B) n (A B)

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Experiment : Roll a pair of dice. (one red, one white)

How many outcomes are there?

How many outcomes are there with no doubles (differentvalue on the two dice) ?

Multiplication Principle Example

6 outcomes on each die. 6 x 6 = 36 outcomes

Think of rolling the dice in succession, say the red f irst.

The red die can take any of 6 outcomes,The white die can take any of 5 value except the one on red.

So there are 6 x 5 ways = 30 outcomes

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Exercise

How many even 4-digit numbers greater than 8000 can be

formed using the digits 1, 2, 3, 6, 7 and 9 if each digit can beused at most once ?

Solution:

The first digit must be 9 so the number is bigger than 8000.

The fourth digit must be 2 or 6 only two choices.

The third digit can be any of 4 choices.

The second digit can be any of 3 choices.

So there are 1x3x4x2 = 24 such numbers

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More Multiplication Exercises

1. A multiple choice exam has 10 questions with 4 choices each.In how many ways can the exam be answered if each question isanswered with one of the choices ?

Answer : 4 4 4 4 4 4 4 4 4 4 = 410 ways

3. How many subsets does a set of 10 elements have ?

Answer: For each subset, an element of the set may ormay not be included.So there are 2 2 2 2 2 2 2 2 2 2 = 210 ways.

2. How many bit strings are there with 10 bits ?

Answer: Each bit may be 0 or 1.

So there are 2 2 2 2 2 2 2 2 2 2 = 210 ways.


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Addition and Multiplication Example

There are 5 different computer books, 6 different mathbooks, and 8 different literature books.How many ways are there to p ick an (unordered) pair

of two books not in the same category ?

Solution:

If computer and math are chosen, 5x6 = 30 waysIf computer and literature, 5x8 = 40 ways

If math and literature, 6x8 = 48 ways.

These 3 types of selection are disjoint,

So there are 30 + 40 + 48 = 118 ways in all.

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Permutations

Permutation is the arrangement of all or part

of a set of objects into distinguishable

sequences.

Each unique ordering is called a permutation.

The ordering of the objects matters !!!

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Permutation Example

How many different three digit numbers can

we make from 7, 8, and 9 if we use eachnumber only once?

We have three numbers to choose from to place inthe hundreds place.After this, we have two left to put in the tens place.Lastly we have one number left over to put in theones place.

3 2 1 = 6 ways

These numbers are:

987 978 897 879 789 798

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Factorial

Factorial : The operation of repeatedly multiplying a

number times the next lower integer until reaching 1.

n! = n (n-1) (n-2) 1

For convenience, define 0! = 1

In previous example, the number of ways is 3! = 6.

How many three-digit numbers with distinct digits can be

formed using the digits 6,7,8,9 and 0 ?

Answer :

5 4 3 = 60

Calculator exercise on Factorials


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Permutation Formula

The number of permutations of r objects from n distinct objects

In earlier example, P(5,3) = 5! / 2!

= (5 4 3 2 1) / (2 1)= 5 4 3 = 60

)!(

!),(

rn

nPrnPrn

==

Example : How many ways are there to elect the president,

vice president, secretary and treasurer of an organizationwith 50 members if all elected officers are different people?

Answer : P(50,4) = 50 49 48 47 = 5,527,200 ways

Calculator exercise on Permutations

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Permutations in a Circle

How many ways are there to seat 12 guests in acircular table with 12 equally spaced chairs ?

Assume 2 circular permutations are the same if one is aresult of a rotation of the other.

Solution :

Fix one guest in a chair and arrange the other in

11! ways.

The number of permuations of n distinct objects

arranged in a circle is ( n 1 ) !

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Combinations

Combination is the number of ways ofselecting r objects from n objects without

regard to ordering

Each unique selection is called a

combination.

Ordering of the selection does NOT matter !!!

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Combination ExampleExample: 5 bowlers must enter a tournament with 3 primary

players and 2 alternates. How many possible teamcombinations could there be?

That is, how many ways can 3 players be selected from 5?Selecting the 3 automatically selects the 2 alternates.

Solution :

The first player can be chosen in 5 ways, the secondin 4 ways, and the third in 3 ways = 5 4 3 = 60 ways.

But the ordering is not important.After the 3 have been chosen, we must remove theordering of 3 players.There are 3 2 1 = 6 such orderings.

So there are 60 / 6 = 10 ways to field the team into the

tournament.


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Combination Formula

The number of permutations of r objects from n distinct objects

)!(!

!),(

rnr

n

r

nCrnC

rn

=

==

In earlier example, C(5,3) = 5! / 3! 2!

= (5 4 3 2 1) / ((3 2 1)(2 1))= 5 4 / 2 1 = 10

Calculator exercise on Combinations

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Permutation : Not all objects distinct

The number of distinct permutations of n things of whichn1 are of a kind, n2 of a second kind, , nk of the kth kind

where n= n1+ n2++ nk is

!!!

!

2121 kk nnn

n

nnn

n

LL

=

This extends the combination formulafrom 2 groups (selected and unselected) to k groups.

How many ways to arrange the letters in the wordNAMAMANATA ?

Answer: 10!/(2!5!2!1!) = (10 9 8 7 6)/(2 2) = 7560 ways

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Exercise

How many ways can 3 red, 4 yellow and 2 blue

bulbs be arranged in a string of Christmas lightswith 9 sockets?

Solution : 9! / (3! 4! 2!)= 9 8 7 6 5 / 3 2 2 = 3 2 7 6 5

= 1260 ways

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rnCrnC

r

n

rnr

n

rnr

n==

=

=

),(

)!(!

!

)(,

Special Case

)!(!

!

)(, rnr

n

rnr

n

=

!!!

!

2121 kk nnn

n

nnn

n

LL

=

If k=2,


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A Combination Application

XXXXXX

SodaJuiceCoffee

Select r objects (repetition allowed) from n types of objects

282

78

6

81=

=

=

+

r

nr

Example: How many different orders of 6 drinks if thereare 3 kinds of drinks? (Coffee, Juice, Soda)

Question: How many domino tiles are there?Each tile is an unordered pair of 0 to 6 dots with doubles allowed.

Answer: 2 selections from 7 objects = C(2+7-1,2)=C(8,2)=28

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Exercises

1. How many ways are there to assign 100 different diplomats to

five different continents?

2. How many ways are there to assign 100 different diplomats tofive different continents if to each continent 20 diplomats must beassigned ?

3. How many ways can 100 identical diplomatic letters besent to five different continents ?

Each diplomat has 5 ways to be assigned, so 5100 ways

Arrange the diplomats in a row and distribute 20 identical briefcases for each continent.

So 100!/(20!)5 ways.

There are 100 orders from 5 different items,So C(100+5-1,100)=C(104,100)

=104 103 102 101 / 4 3 2 1 = 4,598,126

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Review

Combination : n! / ( r! (n r )! )

Order is not important.

Count all orderings of the same elements as 1 only

Permutation : n! / ( n r)!

Order is important.

Count each ordering separately.

With replacement : n * n *

Next selection is unaffected by previous selection.

Without replacement : n * (n 1) *

Previously selected elements

can not be selected in future selections.

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End

Math 103 04 Combinatorics

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