unit vi discrete structures permutations and combinations se (comp.engg.)

Unit VI Discrete Structures

Permutations and Combinations

SE (Comp.Engg.)

• BOTH

• PERMUTATIONS AND COMBINATIONS

• USE A COUNTING METHOD CALLED

FACTORIAL

• Lets start with a simple example.

• A student is to roll a die and flip a coin. How many possible outcomes will there be?

• 1H 2H 3H 4H 5H 6H• 1T 2T 3T 4T 5T 6T

• The number of ways to arrange the letters ABC:

____ ____ ____

Number of choices for first blank? 3 ____ ____

3 2 ___Number of choices for second blank?

Number of choices for third blank? 3 2 1

Permutations

• A Permutation is an arrangement of items in a particular order.

The number of Permutations of n items chosen r at a time, is given by the formula

. 0 where nrrn

nrpn

)!(

!

• Arrange 2 alphabets from a,b,c.

• ab, ba, ac,ca,bc,cb

• 3P2=3!/(3-2)!=6

• A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

2436028*29*30)!330(

!30330

27!

30! p

CIRCULAR PERMUTATIONS

When items are in a circular format, to find the number of different arrangements, divide:

n! / n

• Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?

• 6! 6 = 120

Combination

• A Combination is an arrangement of items in which order does not matter.

. 0 where nrrnr

nrCn

)!(!

!

The number of Combinations of n items chosen r at a time, is given by the formula

• To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?

960,598,21*2*3*4*5

48*49*50*51*52

)!552(!5

!52552

5!47!

52! C

• A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?

101*2

4*5

)!35(!3

!535

3!2!

5! C

Product and Sum Rules

• Product Rule: • If we need to perform procedure 1 AND

procedure 2.There are n1 ways to perform procedure 1 and n2

ways to perform procedure 2.

• There are n1•n2 ways to perform procedure 1 AND procedure 2.

• Sum Rule: If need to perform either procedure 1 OR

procedure 2. There are n1 ways to perform

procedure 1 and n2 ways to perform procedure 2.

• There are n1+n2 ways to perform procedure 1 OR procedure 2.

• This “OR” is an “exclusive OR.” One choice or the other, but not both.

• How many vehicle number plates can be made if each plate contains two different letters followed by three different digits

Two different letters are made in 26P2 ways.

3 different digits are combined in 10P3ways

Total no. of number plates= 26P2 * 10P3

=26!*10!/(24!*7!)

=26*25*10*9*8

=468000

• An 8 member team is to be formed from a group of 10 men and 15 women. In how many ways can the team be chosen if :

(i) The team must contain 4 men and 4 women(ii) There must be more men than women(iii) There must be at least two men

The team must contain 4 men and 4 women = 286650

(ii) There must be more men than women = (iii) There must be at least two men

• Passwords consist of character strings of 6 to 8 characters. Each character is an upper case letter or a digit. Each password must contain at least one digit.

• How many passwords are possible?

• Total number is• # passwords with 6 char. + # passwords

with 7 char. + # pws 8 char.• (=P6+P7+P8).

• P6: # possibilities without constraint : 366

• # exclusions is # passwords without any digits is 266

• And so, P6 = 366-266

• Similarly, P7 = 367-267 and P8 = 368-268

• P = P6+P7+P8 = 366-266 + 367-267 + 368-2

• How many bit-strings of length 8 either begin with 1 or end with 00?

• A = 8-bit strings starting with 1• |A| = # of 8-bit strings starting with 1 is 27

• B = 8-bit strings starting with 00• |B| = # of 8-bit strings ending with 00 is 26

• # of bit-strings begin with 1 and end with 00 is 25.

• # of 8-bit strings starting with 1 or ending with 00 is

• 27+ 26- 25

• |AB| = |A| + |B| - |AB|• Inclusion–Exclusion Principle

33

Permutations with non-distinguishable objects

• The number of different permutations of n objects, where there are non-distinguishable objects of type 1, non-distinguishable objects of type 2, …, and non-distinguishable objects of type k, is

i.e., C(n, )C(n- , )…C(n- - -…- , )

1 2

!! !... !kn

n n n

1n

2nkn

1n 1n 2n 1n 2n 1kn kn

1 2 ... kn n n n

• How many different strings can be made by reordering the letters of the word

OFF

• 3!/2!=3• OFF• FFO• FOF

• ONE• OEN• NEO• NOE• ENO• EON

• How many different strings can be made by reordering the letters of the word

SUCCESS

Generating Permutations

Lexicographic method

• For the following 4• combinations from the set f= {1;2;3;4;5;6;7}

find the combination that immediately follows them in lexicographic order

1234 is followed by3467 is followed by4567 is followed by

What is probability?

• Probability is the measure of how likely an event or outcome is.

• Different events have different probabilities!

How do we describe probability?

• You can describe the probability of an event with the following terms:– certain (the event is definitely going to happen)– likely (the event will probably happen, but not definitely)– unlikely (the event will probably not happen, but it might)– impossible (the event is definitely not going to happen)

• probabilities are expressed as fractions.– The numerator is the number of ways the event

can occur.– The denominator is the number of possible events

that could occur.

6.43

Random Experiment…• …a random experiment is an action or process

that leads to one of several possible outcomes. For example:

Experiment Outcomes

Flip a coin Heads, Tails

Exam Marks Numbers: 0, 1, 2, ..., 100

Assembly Time t > 0 seconds

Course Grades F, D, C, B, A, A+

What is the probability that I will choose a red marble?

• In this bag of marbles, there are:– 3 red marbles– 2 white marbles– 1 purple marble– 4 green marbles

Probability example• Sample space: the set of all possible outcomes.

• Probabilities: the likelihood of each of the possible outcomes (always 0 P 1.0).

)(1)(

)(1)(

1)()(

1)(0

EPEP

EPEP

EPEP

EP

6.46

Probabilities…

• List the outcomes of a random experiment…

• This list must be exhaustive, i.e. ALL possible outcomes included.

• Die roll {1,2,3,4,5} Die roll {1,2,3,4,5,6}

• The list must be mutually exclusive, i.e. no two outcomes can occur at the same time:

• Die roll {odd number or even number} • Die roll{ number less than 4 or even number}

• A and B are independent if and only if P(A&B)=P(A)*P(B)

• A and B are mutually exclusive events:P(A or B) = P(A) + P(B)

6.48

Events & Probabilities…

• The probability of an event is the sum of the probabilities of the simple events that constitute the event.

• E.g. (assuming a fair die) S = {1, 2, 3, 4, 5, 6} and• P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

• Then:• P(EVEN) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 =

1/2

Female

Low

Male

e.g. 1. The following table gives data on the type of car, grouped by petrol consumption, owned by 100 people.

42123

73312

HighMedium

One person is selected at random.L is the event “the person owns a low rated car”F is the event “a female is chosen”.

Find (i) P(L) (ii) P(F L) (iii) P(F| L)

100

Total

(i) P(L) =

Solution:

Find (i) P(L) (ii) P(F L) (iii) P(F L)

100

42123Female

73312Male

HighMediumLow

20 20

7

735

100

(ii) P(F L) =23

Total

100

(iii) P(F L) =23

Notice that

P(L) P(F L)35

23

20

7

100

23

So, P(F L) = P(F|L) P(L)

= P(F L)35

5

1

45RF

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue.

Draw a Venn diagram and use it to illustrate the conditional probability formula.

Solution:

15

10

12

P(R F) =

P(F) =P(R F) = 8

845

208

45

20

20

8

P(R F) = P(R|F) P(F) So,

P(R F) P(F) =

2045

45

8

1

1

Let R be the event “ Red flower ” and F be the event “ First packet ”

Bayes’ Rule: derivation

)(

)&()/(

BP

BAPBAP

• Definition:Let A and B be two events with P(B) 0. The conditional probability of A given B is:

BAPBPBAPBP

BAPBPABP

|~~|

||

Example

• Three jars contain colored balls as described in the table below.– One jar is chosen at random and a ball is selected. If the ball is red,

what is the probability that it came from the 2nd jar?

Jar # Red White Blue

1 3 4 12 1 2 33 4 3 2

Example

• We will define the following events:– J1 is the event that first jar is chosen

– J2 is the event that second jar is chosen

– J3 is the event that third jar is chosen– R is the event that a red ball is selected

Example

• The events J1 , J2 , and J3 mutually exclusive – Why?

• You can’t chose two different jars at the same time

• Because of this, our sample space has been divided or partitioned along these three events

Venn Diagram

• Let’s look at the Venn Diagram

Venn Diagram

• All of the red balls are in the first, second, and third jar so their set overlaps all three sets of our partition

Finding Probabilities

• What are the probabilities for each of the events in our sample space?

• How do we find them?

BPBAPBAP |

Computing Probabilities

• Similar calculations show:

8

1

3

1

8

3| 111 JPJRPRJP

27

4

3

1

9

4|

18

1

3

1

6

1|

333

222

JPJRPRJP

JPJRPRJP

Venn Diagram

• Updating our Venn Diagram with these probabilities:

Where are we going with this?

• Our original problem was:– One jar is chosen at random and a ball is selected.

If the ball is red, what is the probability that it came from the 2nd jar?

• In terms of the events we’ve defined we want:

RPRJP

RJP

22 |

Finding our Probability

RJPRJPRJP

RJP

RP

RJPRJP

321

2

22 |

17.071

12

274

181

81

181

|321

22

RJPRJPRJP

RJPRJP