10-9 Permutations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation...
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Transcript of 10-9 Permutations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation...
10-9 Permutations
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpFind the number of possible outcomes.
1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna
16
Course 3
10-9Permutations
Warm UpFind the number of possible outcomes.
2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham
12
Course 3
10-9Permutations
Warm UpFind the number of possible outcomes.
3. How many different 4–digit phone extensions are possible?
10,000
Course 3
10-9Permutations
Problem of the Day
What is the probability that a 2-digit whole number will contain exactly one 1?
Course 3
10-9Permutations
1790
Learn to find permutations.
Course 3
10-9Permutations
Vocabulary
factorialpermutation
Insert Lesson Title Here
Course 3
10-9Permutations
Course 3
10-9Permutations and Combinations
The factorial of a number is the product of all the whole numbers from the number down to 1. The factorial of 0 is defined to be 1.
5!5! = 55 • 44 • 33 • 22 • 11
Read 5! as “five factorial.”
Reading Math
Evaluate each expression.
Example 1: Evaluating Expressions Containing Factorials
Course 3
10-9Permutations
A. 9!
9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880
8!6!
8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1
Write out each factorial and simplify.
8 • 7 = 56
B.
Multiply remaining factors.
Example 1: Evaluating Expressions Containing Factorials
Course 3
10-9Permutations and Combinations
Subtract within parentheses.
10 • 9 • 8 = 720
10!7!
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7 6 5 4 3 2 1
C. 10!
(9 – 2)!
Evaluate each expression.
Check It Out: Example 1
Course 3
10-9Permutations
A. 10!
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 3,628,800
7!5!
7 • 6 • 5 • 4 • 3 • 2 • 1 5 • 4 • 3 • 2 • 1
Write out each factorial and simplify.
7 • 6 = 42
B.
Multiply remaining factors.
Check It Out: Example 1
Course 3
10-9Permutations
Subtract within parentheses.
9 • 8 • 7 = 504
9!6!
9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 6 5 4 3 2 1
C. 9!
(8 – 2)!
Course 3
10-9Permutations
A permutation is an arrangement of things in a certain order.
If no letter can be used more than once, there are 6 permutations of the first 3 letters of the alphabet: ABC, ACB, BAC, BCA, CAB, and CBA.
first letter
?
second letter
?
third letter
?
3 choices 2 choices 1 choice
The product can be written as a factorial.
• •
3 • 2 • 1 = 3! = 6
Course 3
10-9Permutations
If no letter can be used more than once, there are 60 permutations (orders) of the first 5 letters of the alphabet, when taken 3 at a time: ABE, ACD, ACE, ADB, ADC, ADE, and so on.
first letter
?
second letter
?
third letter
?
5 choices 4 choices 3 choices
5 • 4 • 3 = 60 permutations
Course 3
10-9Permutations
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
ACB ADB AEB ADC AEC AED BDC BEC BED CED
BAC BAD BAE CAD CAE DAE CBD CBE DBE DCE
BCA BDA BEA CDA CEA DEA DBC CEB DEB DEC
CAB DAB EAB DAC EAC EAD DCB EBC EBD ECD
CBA DBA EBA DCA ECA EDA DBC ECB EDB EDC
These 6 permutations are all the same combination.
In the list of 60 permutations, each combination is repeated 6 times. The number of combinations is = 10.
60 6
Jim has 6 different books.
Example 2A: Finding Permutations
Course 3
10-9Permutations
Find the number of orders in which the 6 books can be arranged on a shelf.
7206 • 5 • 4 • 3 • 2 • 1 =
There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.
Course 3
10-9Permutations
= 5040 7! = 7 • 6 • 5 • 4 • 3 • 2 • 1
Use 7!
There are 5040 orders in which to arrange 7 soup cans.
Check It Out: Example 2A
Find the number of orders in which all 7 soup cans can be arranged on a shelf.
There are 7 soup cans in the pantry.
Example 2B: Finding Permutations
Course 3
10-9Permutations
If the shelf has room for only 3 of the books, find the number of ways 3 of the 6 books can be arranged.
There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.
6 • 5 • 4 =6P3
The number of books is 6.
The books are arranged 3 at a time. = 120
Course 3
10-9Permutations
There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.
= 7 • 6 • 5 • 47P4
The number of cans is 7.
The cans are arranged 4 at a time. = 840
There are 7 soup cans in the pantry.Check It Out: Example 2B
If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.
Evaluate each expression.
1. 9!
2.
3. There are 8 hot air balloons in a race. In how many possible orders can all 8 hot air balloons finish the race?
Lesson Quiz
3024
362,880
Insert Lesson Title Here
40,320
Course 3
10-9Permutations
9!5!