1. Part 2 – Factorial and other Counting Rules
Probability
Part 2 – Factorial and other Counting Rules

2. 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?

3. Probability Agenda Warm-up Homework Review
Objective: To understand and apply concepts related to the factorial counting rule, permutations, and combinations.
Summary
Homework

4. 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.

5. 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

6. 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

7. 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)

8. Permutations
Consider the possible arrangements of the letters a, b, and c.
The possible arrangements are: abc, acb, bac, bca, cab, cba.
If the order 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.

9. Permutations
An arrangement of n distinct objects in a specific order is called a permutation of the objects.
Note: To determine the number of possibilities mathematically, one can use the multiplication rule to get: 3  2  1 = 6 permutations.

10. Permutations
Permutation Rule : The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taken r objects at a time. It is written as nPr and the formula is given by nPr = n! / (n – r)!.

11. 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: Number of ways = 7P2 = 7! / (7 – 2)! = 7!/5! = 42.

12. 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: Number of ways =9P4 = 9! / (9 – 4)! = 9!/5! = 3024.

13. Combinations
Consider the possible arrangements of the letters a, b, and c.
The possible arrangements are: abc, acb, bac, bca, cab, cba.
If the order 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.

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

15. Combinations - Example
How many combinations of four objects are there taken two at a time?
Solution: Number of combinations:
4C2 = 4! / [(4 – 2)! 2!] = 4!/[2!2!] = 6.

16. 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: Number of combinations:
12C5 = 12! / [(12 – 5)! 5!] = 12!/[7!5!] = 792.

17. 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: Number of possibilities: (number of ways of selecting 3 women from 7) (number of ways of selecting 2 men from 5) = 7C3 5C2 = (35)(10) = 350.

18. 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?

19. Combinations - Example
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: C3 5C2 + 8C4 5C1 + 8C5 5C0= (56)(10) + (70)(5) + (56)(1) = 966.

20. 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

21. Probability
Summary
Factorial Counting Rule
Permutations
Combinations

22. Probability
Homework
Probability Practice Sheet 2