Counting Principles (Permutations and Combinations )

Counting Principles(Permutations and Combinations )

© 2012 Pearson Education, Inc.. All rights reserved.

Example

A combination lock can be set to open to any 4-digit

sequence.

(a) How many sequences are possible?

(b) How many sequences are possible if no digit is repeated?


Solution: (a) Since there are 10 digits namely 0, 1, 2…..9, thereare 10 choices for each of the digit. By the multiplicationprinciple, there are 10 ∙10 ∙10 ∙10 =10,000 different sequences.

(b)There are 10 choices for the first digit. It cannot be usedagain, so there are 9 choices for the second digit, 8 choices forthe third digit, and then 7 choices for the fourth digit.Consequently, the number of such sequences is 10 ∙ 9 ∙ 8 ∙7 =5040 different sequences.


Example

How many different ways can you choose a bagel, muffin or donut to eat and coffee or juice to drink?

Tree diagram

Example

A teacher is lining up 8 students for a spelling bee. How many

different line-ups are possible?


Solution: Eight choices will be made, one for each space that willhold a student. Any of the students could be chosen for the first space. There are 7 choices for the second space, since 1 studenthas already been placed in the first space; there are 6 choices forthe third space, and so on. By the multiplication principle, the number of different possiblearrangements is 8 ∙7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙1 = 40,320.


TI-83/84 function

(order does matter)

Example

A teacher wishes to place 5 out of 8 different books on her

shelf. How many arrangements of 5 books are possible?


Solution : The teacher has 8 ways to fill the first space, 7 waysto fill the second space, 6 ways to fill the third, and so on…Since the teacher wants to use only 5 books, only 5 spaces canbe filled (5 events) instead of 8, for 8 ∙7 ∙ 6 ∙ 5 ∙ 4 = 6720arrangements.

Example

Find the number of permutations of the letters L, M, N, O, P,

and Q, if just three of the letters are to be used.

Solution:


!( , ) . Here 6 and 3.

( - )!n

P n r n rn r

6! 6!(6 -3)! 3!

6 5 4 3 2 1120

3 2 1


How many permutations are there of two letters from the set {A,B,C}.

)!(

!),(

rn

nrnP

623!1

!3

)!23(

!3)2,3(

P


(order does not matter)

TI-83/84 function

Example

How many committees of 4 people can be formed

from a group of 10 people?

Solution: A committee is an unordered group, so use the

combinations formula for C(10,4).


10! 10!(10,4)

4!(10 4)! 4!6!C

10 9 8 7 6 5 4 3 2 14 3 2 1 6 5 4 3 2 1

10 9 8 74 3 2 1

210


Example

From a class of 15 students, a group of 3 or 4 students will be

selected to work on a special project. In how many ways can a

group of 3 or 4 students be selected?

Solution: The number of ways to select group of 3 students

from a class of 15 students is C(15, 3) = 455.

The number of ways to select group of 4 students from a class

of 15 students is C(15, 4) = 1365.

The total number of ways to select a group of 3 or 4 students

will be the 1820.



Example

(a) How many 4-digit code numbers are possible if no digits are repeated?

Solution: Since changing the order of the 4 digits results in a

different code, permutations should be used.

(b) A sample of 3 light bulbs is randomly selected from a batch

of 15. How many different samples are possible?

Solution: The order in which the 3 light bulbs are selected is not

important. The sample is unchanged if the items are rearranged,

so combinations should be used.


Example

(c) In a baseball conference with 8 teams, how many games

must be played so that each team plays every other team exactly once?

Solution: Selection of 2 teams for a game is an unordered

subset of 2 from the set of 8 teams. Use combinations again.

(d) In how many ways can 4 patients be assigned to 6 different

hospital rooms so that each patient has a private room?

Solution: The room assignments are an ordered selection of 4

rooms from the 6 rooms. Exchanging the rooms of any 2

patients within a selection of 4 rooms gives a different

assignment, so permutations should be used.


Introduction to Probability

Write the elements belonging to the set {x | x is a state whose

name begins with the letter O}.


Solution: The state names that begin with the letter O make upthe set {Ohio, Oklahoma, Oregon }.

Set – a collection of elements that satisfy a certain condition.


Example 1

Decide if the statement is true or false.

Solution: The first set is a subset of the second because each element of first set belongs to second set. (The fact that the elements are listed in a different order does not matter.) Therefore, the statement is true.

{2,4,6} {6,2,4}



A) What is the sample space?

B) What is the probability of getting a 3?

C) What is the probability of getting an odd number?

Sample Space – The set of all possible outcomes of an experiment or event.

Example

Two coins are tossed, and a head or a tail is recorded for each

coin. Write the event E: the coins show exactly one head.

Solution: Tossing a coin is made up of the outcomes heads (H)

or tails (T). If S represents the sample space of tossing two

coins, then S = {HH, HT, TH, TT}.

Two outcomes satisfy this condition: so,

E = {HT, TH}.


ExampleTwo coins are tossed, and a head or a tail is recorded for each

coin.

A) Give a sample space for this experiment.

B) What is the probability of getting at least one head.

Solution: A) Tossing a coin is made up of the outcomes heads (H)

or tails (T). If S represents the sample space of tossing two

coins then S = {HH, HT, TH, TT}.

Solution : B) E = {HH,HT,TH}


Number of elements in the sets

4

3

)(

)()(

Sn

EnEP

384.0 152.0 824.056.0

616.0 712.0

179.0

0625.0

625.025.0

75.0

http://www.youtube.com/watch?v=mhlc7peGlGg&safe=active

Monty Hall probability puzzle




Probability of Multiple Events


46

5

1

2

6FE

(no outcomes in common; if one event occurs the other cannot)

(E’ is called the complement of E)

Example

A study of workers earning the minimum wage grouped such

workers into various categories, which can be interpreted as

events when a worker is selected at random. Source:

Economic Policy Institute. Consider the following events:

E: worker is under 20;

F: worker is white;

G: worker is female.

Describe the following event in words:

Solution:


.E F

is the event that the worker is not under 20

and the worker is not white.

E F


Independent Events

The fact that one event has occurred, does not change the probability of the second event occurring.

** Events that are mutually exclusive must be dependent, but dependent events are not necessarily mutually exclusive**

Read as; the probability of event F occurring given E has occurred

Read as; the probability of event E occurring given F has occurred


(General rule of addition)

For mutually exclusive events,

So the equation simplifies to

0)( FEP)()()( FPEPFEP

Example If a single card is drawn from an ordinary deck of cards, find

the probability of an ace or a club.

Solution: Let A represent the event “an ace ” and C the event

“club card.” There are 4 aces in the deck, so

There are 13 clubs in the deck, so

Since there is 1 ace of club in the deck,

By the union rule, the probability of the card being an ace or a

Club card is


( ) 4 / 52.P A

( ) 13 / 52.P C

( ) 1/ 52.P A C

( ) ( ) ( ) ( ).P A C P A P C P A C

4 13 152 52 52

16 4.

52 13

ExampleSuppose two fair dice are rolled. Find the probability that the sum is 8, or both

die show the same number.

Solution:

1-1 1-2 1-3 1-4 1-5 1-6

2-1 2-2 2-3 2-4 2-5 2-6

3-1 3-2 3-3 3-4 3-5 3-6

4-1 4-2 4-3 4-4 4-5 4-6

5-1 5-2 5-3 5-4 5-5 5-6

6-1 6-2 6-3 6-4 6-5 6-6


By the union rule, (sum is 8 or both die show same number)

5 6 1 10 5= .

36 36 36 36 18

P

Example

Find the probability that when two fair dice are rolled, the sum

is less than 11.

Solution: To calculate this probability directly, we must find the

probabilities that the sum is 2, 3, 4, 5, 6, 7, 8, 9 or 10 and then

add them. It is much simpler to first find the probability of the

complement, the event that the sum is greater than or equal to 11.

1-1 1-2 1-3 1-4 1-5 1-6

2-1 2-2 2-3 2-4 2-5 2-6

3-1 3-2 3-3 3-4 3-5 3-6

4-1 4-2 4-3 4-4 4-5 4-6

5-1 5-2 5-3 5-4 5-5 5-6

6-1 6-2 6-3 6-4 6-5 6-6© 2012 Pearson Education, Inc.. All rights reserved.

(sum 11) 1 (sum 11)P P

3 33 111 .

36 36 12

(General rule of multiplication)

When events are Independent;

So the equation simplifies to:

)()|( EPFEP )()|( FPEFP

Independent

Not Independent

Independent

Not Independent

1/6

0.54 2/7

9/34

9/25

Not mutually exclusive

mutually exclusive

Not mutually exclusive


P(F); represented by large circle.

P(E and F); represented by overlapping section

Example

The following table shows national employment statistics. Use the table to findeach probability.

7. P(male | professional) 8. P(laborer | female) 9. P(female | sales)

10.P(professional | female) 11.P(sales | male) 12.P(male | laborer)

8937 5801 43839392

9729

19121

47.08937

4190

27.09729

2588 67.0

4383

2951

15.09392

1432 55.0

5801

3213

51.09392

4747

ExampleUse a tree diagram to solve the following problem.

16. A car insurance company compiled the following information from a recentsurvey. 75% of drivers carefully follow the speed limit Of the drivers who carefully follow the speed limit, 80% have never had an accident. Of the drivers who do not carefully follow the speed limit, 65% have never had an accident.

What is the probability that a driver does not carefully follow the speed limit and has never had an accident?

15.0)'( NACP0.75

0.65

0.20

0.80

0.25

0.35

C is the event careful driver.

NA is the event no accident

1625.0)'( NACP

6.0)( NACP

0875.0)''( NACP

Documents

Counting Principles (Permutations and Combinations )