Probability

Part 2 – Factorial and other Counting Rules

Probability

Warm-up Canadian “zip” codes are as follows

NLN – LNL, where N is a number and L is a letter. How many “zip” codes are possible?

Probability

Agenda Warm-up Homework Review Objective: To understand and apply concepts

related to the factorial counting rule, permutations, and combinations.

Summary Homework

Factorial Counting Rule

If you have n items that occur in a sequence of events and no repetitions are permitted, then the number of ways the sequence can occur is n!

Where n is the number of items.

Probability

Example- You have five items you wish to arrange items on

a shelf. How many different ways can they be arranged?

5! = 5 x 4 x 3 x 2 x 1 = 120

Partitions Rule

Determines the number of different ways, you can partition the elements of a set of n elements into k groups consisting of n1, n2, …, nk objects respectively

nnnnn

n k

ii

k

121

where!!!

!

Partitions Example

You have 10 children to help around the house. You want to assign three to clean up the yard, four to help paint the downstairs and three to wash the family car. In how many different ways can you group your children? k = 3 (different areas) n1= 3, n2=4, n3=3 (group sizes)

200,43!4!3!

10!

!!!

!

21

knnn

n

Permutations

Consider the possible arrangements of the letters aa, bb, and cc.

The possible arrangements are: abc, acb, bac, abc, acb, bac, bca, cab, cba.bca, cab, cba.

If the order of the arrangement is importantorder of the arrangement is important then we say that each arrangement is a permutation of the three letters. Thus there are six permutations of the three letters.

Permutations

An arrangement of n distinct objects in a specific order is called a permutationpermutation of the objects.

Note:Note: To determine the number of possibilities mathematically, one can use the multiplication rule to get: 3 2 1 = 6 permutations.

Permutations

Permutation Rule :Permutation Rule : The arrangement of nn objects in a specific order using r objects at a time is called a permutation of nn objects taken rr objects at a time. It is written as nnPPrr and the formula is given by

nnPPrr = n! / (n – r)! = n! / (n – r)!.

Permutations - - Example

How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available?

Solution:Solution: Number of ways = 77PP22 = 7! / (7 – 2)! = 7! / (7 – 2)! = 7!/5! = 42= 7!/5! = 42.

Permutations - - Example

How many different ways can four different books be arranged in a specific location on a shelf if they can be selected from nine different books?

Solution:Solution: Number of ways =99PP44 = 9! / (9 – 4)! = 9! / (9 – 4)! = 9!/5! = 3024= 9!/5! = 3024.

Combinations

Consider the possible arrangements of the letters aa, bb, and cc.

The possible arrangements are: abc, acb, bac, abc, acb, bac, bca, cab, cba.bca, cab, cba.

If the order of the arrangement is not importantorder of the arrangement is not important then we say that each arrangement is the same. We say there is one combination of the three letters.

Combinations

Combination Rule :Combination Rule : The number of combinations of rr objects from nn objects is denoted by nnCCrr and the formula is given by nnCCrr = n! / [(n – r)!r!] = n! / [(n – r)!r!] .

Combinations - - Example

How many combinations of four objects are there taken two at a time?

Solution:Solution: Number of combinations:

44CC22 = 4! / [(4 – 2)! = 4! / [(4 – 2)! 2!] = 4!/[2!2!] = 6.2!] = 4!/[2!2!] = 6.

Combinations - - Example

In order to survey the opinions of customers at local malls, a researcher decides to select 5 malls from a total of 12 malls in a specific geographic area. How many different ways can the selection be made?

Solution:Solution: Number of combinations: 1212CC55 = 12! / [(12 – 5)! = 12! / [(12 – 5)! 5!] = 12!/[7!5!] = 792.5!] = 12!/[7!5!] = 792.

Combinations - - Example

In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?

Solution:Solution: Number of possibilities: (number of ways of selecting 3 women from 7) (number of ways of selecting 2 men from 5) = 77CC33 55CC22 = =

(35)(10) = 350.(35)(10) = 350.

Combinations - - Example

A committee of 5 people must be selected from 5 men and 8 women. How many ways can the selection be made if there are at least 3 women on the committee?

Combinations - - Example

Solution:Solution: The committee can consist of 3 women and 2 men, or 4 women and 1 man, or 5 women. To find the different possibilities, find each separately and then add them: 88CC33 55CC22 + + 88CC44 55CC11 + + 88CC55 55CC00= (56)(10) + (70)(5) + = (56)(10) + (70)(5) + (56)(1) = 966.(56)(1) = 966.

Probability – Counting Rules

1) A television news director wishes to use three news stories on an evening show. One story will be the lead story, one will be the second story, and the last will be a closing story. If the director has a total of eight stories to choose from, how many possible ways can the program be set up? 336

2) How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available? 42

3) A bicycle shop owner has 12 mountain bicycles in the showroom. The owner wishes to select 5 of them to display at a bicycle show. How many different ways can a group of 5 be selected? 792

4) In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there? 350

Probability

Summary Factorial Counting Rule Permutations Combinations

Probability

Homework Probability Practice Sheet 2