copyright © 1998, triola, elementary statistics addison wesley longman 1 counting section 3-7 m a r...

Copyright © 1998, Triola, Elementary Statistics

Addison Wesley Longman1

CountingCountingSection 3-7Section 3-7

M A R I O F. T R I O L ACopyright © 1998, Triola, Elementary Statistics

Addison Wesley Longman



Fundamental Counting Rule

For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m • n ways.



a

b

c

d

e

a

b

c

d

e

T

F

Tree Diagram of the events

T & aT & bT & cT & dT & eF & aF & bF & cF & dF & e



a

b

c

d

e

a

b

c

d

e

T

F

Tree Diagram of the events

T & aT & bT & cT & dT & eF & aF & bF & cF & dF & e

m = 2 n = 5 m*n = 10



Rank three players (A, B, C). How many possible outcomes are there?

Ranking: First Second Third

Number of Choices: 3 2 1

By FCR, the total number of possible outcomes are:

3 * 2 * 1 = 6

( Notation: 3! = 3*2*1 )

Example



The factorial symbol ! denotes the product of decreasing positive whole numbers.

n! = n (n – 1) (n – 2) (n – 3) • • • • • (3) (2) (1)

Special Definition: 0! = 1

Find the ! key on your calculator

Notation



A collection of n different items can be arranged in order n! different ways.

Factorial Rule



Eight players are in a competition, three of them will win prices (gold/silver/bronze). How many possible outcomes are there?

Prices: gold silver bronze

Number of Choices: 8 7 6

By FLR, the total number of possible outcomes are:

8 * 7 * 6 = 336 = 8! / 5!

= 8!/(8-3)!

Example



n is the number of available items (none identical to each other)

r is the number of items to be selected

the number of permutations (or sequences) is

Permutations Rule



n is the number of available items (none identical to each other)

r is the number of items to be selected

the number of permutations (or sequences) is

Permutations Rule

Order is taken into account

Pn r = (n – r)!n!



when some items are identical to others

Permutations Rule




If there are n items with n1 alike, n2 alike, . . .

nk alike, the number of permutations is

Permutations Rule




If there are n items with n1 alike, n2 alike, . . .

nk alike, the number of permutations is

Permutations Rule

n1! . n2! .. . . . . . . nk! n!



Eight players are in a competition, top three will be selected for the next round (order does not matter). How many possible choices are there?

By the Permutations Rule, the number of choices of Top 3 with order are 8!/(8-3)!

For each chosen top three, if we rank/order them, there are 3! possibilities.

==> the number of choices of Top 3 without order are {8!/(8-3)!}/(3!)

Example

8!

(8-3)! 3!=



Eight players are in a competition, top three will be selected for the next round (order does not matter). How many possible choices are there?

By the Permutations Rule, the number of choices of Top 3 with order are 8!/(8-3)!

For each chosen top three, if we rank/order them, there are 3! possibilities.

==> the number of choices of Top 3 without order are {8!/(8-3)!}/(3!)

Example

8!

(8-3)! 3!= Combinations

rule!



the number of combinations is

Combinations Rule



n different items

r items to be selected

different orders of the same items are not counted

the number of combinations is

(n – r )! r!n!

nCr =

Combinations Rule



When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.



Is there a sequence of events in which the first can occur m ways, the second can

occur n ways, and so on?

If so use the fundamental counting rule and multiply m, n, and so on.

Counting Devices Summary



Are there n different items with all of them to be used in different arrangements?

If so, use the factorial rule and find n!.




Are there n different items with some of them to be used in different arrangements?




Are there n different items with some of them to be used in different arrangements?

If so, evaluate


(n – r )!n!

n Pr =



Are there n items with some of them identical to each other, and there is a need to find the total number of different arrangements of all of those n items?




Are there n items with some of them identical to each other, and there is a need to find the total number of

different arrangements of all of those n items?

If so, use the following expression, in which n1 of the

items are alike, n2 are alike and so on


n!n1! n2!. . . . . . nk!



Are there n different items with some of them to be selected, and is there a need to find the total number of combinations (that is, is the order irrelevant)?




Are there n different items with some of them to be selected, and is there a need to find the total number of combinations (that is, is the order irrelevant)?

If so, evaluate


n!nCr = (n – r )! r!

copyright © 1998, triola, elementary statistics addison wesley longman 1 counting section 3-7 m a r...

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