# Decisions, Decisions, Decisions

## Presentation on theme: "Decisions, Decisions, Decisions"— Presentation transcript:

Decisions, Decisions, Decisions
Permutations and Combinations Decisions, Decisions, Decisions

Objectives I will : Compare and contrast permutations and combinations. Understand terminology and variables associated with permutations and combinations. Identify and apply permutations and combinations in problem situations. Use technology to compute probabilities of permutations and combinations.

First A review of the Multiplication Rule
If there are n possible outcomes for event A, and m possible outcomes for event B, then there are n x m possible outcomes for event A followed by event B. # of Outcomes = n x m .

Toss Coin, Roll Dice Example
Head 1 2 3 4 5 6 Tail Roll Dice Toss Coin, Roll Dice Example Resulting Sample Space H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6

Multiplication Rule If there are n possible outcomes for event A and m possible outcomes for event B, then there are n x m possible outcomes for event A followed by event B. In our Coin Toss, Die Roll example, there are 2 possible outcomes for the Coin Roll (H or T) and 6 possible events for the Die Toss (1,2,3,4,5, or 6). There are 2 x 6 or 12 possible outcomes in the sample space.

Multiplication Rule More Examples
Ford’s new Fusion comes with two body styles, three interior package options, and four colors, as well as a choice of standard or automatic transmissions. The dealership wants to carry one of each type in its inventory. How many Ford Fusions will they order? There are 2 body styles, 3 interior package options, 4 colors, and 2 transmissions. There are 2 x 3 x 4 x 2 = 48 possible outcomes.

Multiplication Rule More Examples
The Old Orange Café offers a special lunch menu each day including two appetizers, three main courses, and four desserts. Customers can choose one dish from each category. How many different meals can be ordered from the lunch menu? There are 2 appetizers, 3 main courses, and 4 desserts. There are 2 x 3 x 4 = 24 possible outcomes.

ReVIEW Factorial Notation
Remember, for a counting number, n n! = n (n-1) (n-2) (n-3) …1 i.e., 5! = 5 x 4 x 3 x 2 x 1 0! = 1 (By special definition) 1! = 1

Combinations The number of ways choices can be combined without repetition. Order does not matter. For example, how many ways can these four members of the tennis team be combined to play a doubles match?

Combinations How many ways can these four members of the tennis team be combined to play a doubles match? The multiplication rule says there are 4 possible choices for the 1st team member x 3 possible choices for the 2nd team member. Or 4 x 3 = 12 possible choices. However, a quick illustration shows this is incorrect. There are in fact only 6 possible unique combinations. 1 3 4 5 6 2

Counting Rule for Combinations
The number of combinations of n distinct objects , taking them r at a time, is nCr or Cn,r = n! r!(n-r)! Where n and r are whole numbers. n ≥ r and n! (“n factorial”) = n x (n – 1) x (n – 2) x …. 1

Combinations Using the combination formula nCr or Cn,r in our tennis team example, we calculate: The number of combinations of 4 distinct players , taken 2 at a time, is 4C2 or C4,2 = 4! = 4!___ = 4 x 3 x 2 x 1 = 6 2!(4-2)! !(2)! x 1 x 2 x 1 1 3 4 5 6 2

Consider another Sample Space
How many ways can these five members of the tennis team be combined to play a doubles match? Using the combination formula nCr or Cn,r we calculate: The number of combinations of 5 distinct players , taken 2 at a time, is 5C2 or C5,2 = 5! = 5!___ = 5 x 4 x 3 x 2 x 1 = 10 2!(5 - 2)! !(3)! x 1 x 3 x 2 x 1

And another Sample Space
How many ways can these six members of the track team be combined to run a 4 x 4 relay? Using the combination formula nCr or Cn,r we calculate: The number of combinations of 6 distinct runners , taken 4 at a time, is 6C4 or C6,4 = 6! = 6!___ = 6 x 5 x 4 x 3 x 2 x 1 = 15 4!(6 - 4)! !(2)! x 3 x 2 x 1 x 2 x 1

Ordered arrangements are Permutations
Sometimes we need to consider how many different ways n items can be ordered. For example, how many different ways can 8 people be seated around a table? For the first seat there are 8 choices, for the second seat there are 7 choices, for the third seat there are 6 choices, etc. The number of possible ordered arrangements is : 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 or 8! ( “ 8 factorial) . Ordered arrangements are Permutations

Permutations The number of possible ordered arrangements is
How many different ways can 10 sprinters finish an Olympic trial? For 1st place there are 10 choices, for 2nd place there are 9 choices, for 3rd place there are 8 choices, for 4th place there are 7 choices, for 5th place there are 6 choices, for 6th place there are 5 choices, for 7th place there are 4 choices, for 8th place there are 3 choices, for 9th place there are 2 choices, and for the 10th and final place there is only one remaining choice. The number of possible ordered arrangements is 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 of 10! (“10 factorial”)

Consider another Sample Space
How many ways can three members of the track team finish a race in the top three positions? 5 1 3 6 2 4 A simple illustration shows there are just 6 possible combinations.

Counting Rule for Permutations
But what happens when only some of the objects will be chosen? The number of ways to arrange in order n distinct objects , taking them r at a time, is nPr or Pn,r = n! (n-r)! Where n and r are whole numbers. n ≥ r and n! (“n factorial”) = n x (n – 1) x (n – 2) x …. 1

Let’s Take Another look at the Sample Space
How many ways can three members of the track team finish a race in the top three positions? Using the permutation formula nPr or Pn,r we calculate: The number of permutations of 3 distinct players , taken 3 at a time, is 3P3 or P3,3 = 3! = 3!___ = 3 x 2 x 1 = 6 (3 - 3)! !

Consider another Sample Space
How many ways can these five members of the track team finish a race in the top three positions? Using the permutation formula nPr or Pn,r we calculate: The number of permutations of 5 distinct players , taken 3 at a time, is 5P3 or P5,3 = 5! = 5!___ = 5 x 4 x 3 x 2 x 1 = 60 (5 - 3)! ! x 1

And another Sample Space
How many ways can these six members of the track team finish a race in the top 3 positions? Using the permutation formula nPr or Pn,r in our track team example, we calculate: The number of permutations of 6 distinct players , taken 3 at a time, is 6P3 or P6,3 = 6! = 6!___ = 6 x 5 x 4 x 3 x 2 x 1 = 120 (6 - 3)! ! x 2 x 1

Combinations and Permutations
No Order nCr or Cn,r = n! r!(n-r)! Permutations 1st 2nd 3rd Order Matters nPr or Pn,r = n! (n-r)!

Credits Designer : Cynthia Toliver Co- Stars : Combinations, nCr Permutations, nPr Supporting Characters Boy Bro Wry Con DJ Dude