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![Page 1: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/1.jpg)
Statistics and Modelling Course
2011
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ARRANGEMENTS&
SELECTIONS
Part of Achievement Standard 90643 - Probability
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WHAT WILL I LEARN FROM THIS UNIT?
Answer: To figure out the number of possible ways in which things can happen.At the end of this unit, you’ll be able to do this for several different types of scenarios.
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A group of friends decide to go to the beach during the day, then have a fast-food meal, and then go to a movie complex in the evening. This table shows the different suggested alternatives.
Beach Fast food restaurant Movie complex
Sumner Burger King Riccarton
Taylor’s Mistake KFC Northlands
New Brighton McDonald’s
Waimairi Beach Dominoes
Subway
For example, they could go to New Brighton, then eat at Dominoes and go to the movies at Westfield Riccarton. In how many different ways could the group of friends spend their day?
Draw a diagram showing the different choices possible.
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If they go to Sumner:
Fast food restaurant
B
K
M
D
S
Beach
S
T
N
W
Movie complex
R
N
They have 5 choices of restaurant.
Sumner Taylor’s New Brighton Waimairi BeachBeach
Restaurant Burger King KFC McDonald’s Dominoes Subway
Movie Riccarton Northlands
For each restaurant, they have 2 choices of movie complex. i.e. 5 2 choices in all.
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Beach Fast food restaurant Movie complex
S
T
N
W
B
K
M
D
S
R
N
They have 5 2 choices choices of restaurant and movie complex.
Sumner Taylor’s New Brighton Waimairi BeachBeach
Restaurant Burger King KFC McDonald’s Dominoes Subway
Movie Riccarton Northlands
But there are 4 beaches to choose from.
If they go to Sumner:
i.e. 4 5 2 = 40 choices in all.
![Page 7: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/7.jpg)
So, in this case, we found that there were 40 possible scenarios for this group of friends’ day out.
Number of Scenarios = Number of beach options
× Number of fast-food
options
× Number of movie options
= 4 × 5 × 2
= 40
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TheMULTIPLICATION PRINCIPLE:
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TheMULTIPLICATION PRINCIPLE:
If one task (e.g. going to a beach) can be performed in n1 ways,
a second task (e.g. getting a feed) can be performed in n2 ways,
and a third task (e.g. going to the movies) can be performed in n3 ways,
then the total number of ways of performing all three tasks = n1 × n2 × n3
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If one task (e.g. going to a beach) can be performed in n1 ways,
a second task (e.g. getting a feed) can be performed in n2 ways,
and a third task (e.g. going to the movies) can be performed in n3 ways,
then the total number of ways of performing all three tasks = n1 × n2 × n3
E.g. 1:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
Show all of the possibilities on a tree diagram.
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The possibilities are shown in the diagram.
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Choose A as the 1st person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
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A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Once the 1st person has been
chosen, there are two ways of
choosing the 2nd person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 13: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/13.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Once the 2nd person has been
chosen, there is only one way of choosing the 3rd person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 14: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/14.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 15: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/15.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Choose B as the 1st person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 16: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/16.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Once the 1st person has been
chosen, there are two ways of
choosing the 2nd person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 17: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/17.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Once the 2nd person has been
chosen, there is only one way of choosing the 3rd person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 18: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/18.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 19: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/19.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Choose C as the 1st person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 20: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/20.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Once the 1st person has been
chosen, there are two ways of
choosing the 2nd person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 21: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/21.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Once the 2nd person has been
chosen, there is only one way of choosing the 3rd person.
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 22: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/22.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 23: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/23.jpg)
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
The possibilities are shown in the diagram.
![Page 24: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/24.jpg)
ABC
ACB
BAC
BCA
CAB
CBA
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
From the diagram, there are 6 ways altogether
E.g. 2:Alex, Barbara and Colin are waiting to enter a cave. In how many different orders can they enter the cave in single file?
![Page 25: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/25.jpg)
ABC
ACB
BAC
BCA
CAB
CBA
A
B
C
B
C
A
C
A
B
C
B
C
A
B
A
Using the Multiplication Principle There are three positions to fill:
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Using the Multiplication Principle
3
The first position can be filled in 3 ways
There are three positions to fill:
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3
Once the first person is chosen, the second person can be chosen in any one of 2 ways as one person is already used
There are three positions to fill:
2
8.01E Alex, Barbara and Colin are waiting to enter a cave. In how many ways can they enter the cave in single file?
Using the Multiplication Principle
![Page 28: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/28.jpg)
3
Once the second person is chosen, the third person can be chosen in only one way
There are three positions to fill:
2 1
8.01E Alex, Barbara and Colin are waiting to enter a cave. In how many ways can they enter the cave in single file?
Using the Multiplication Principle
![Page 29: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/29.jpg)
= 6
3 2 1
8.01E Alex, Barbara and Colin are waiting to enter a cave. In how many ways can they enter the cave in single file?
Using the Multiplication Principle
Number of possible orders =
We can write 3×2×1 as 3! For short.“3!” Is said “3 factorial”.
Factorials
![Page 30: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/30.jpg)
Using the multiplication principle:
= 6
3 2 1 Number of possible orders =
We can write 3×3×2 as 3! For short.“3!” Is said “3 factorial”.
Factorials
![Page 31: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/31.jpg)
Using the multiplication principle:
= 6
3 2 1 Number of possible orders =
We can write 3×3×2 as 3! For short.“3!” Is said “3 factorial”.
FactorialsIn a situation where all of the available
options must be used,
![Page 32: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/32.jpg)
Using the multiplication principle:
= 6
3 2 1 Number of possible orders =
We can write 3×3×2 as 3! For short.“3!” Is said “3 factorial”.
FactorialsIn a situation where all of the available
options must be used, but where each can be used only once (i.e. without replacement),
![Page 33: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/33.jpg)
Using the multiplication principle:
= 6
3 2 1 Number of possible orders =
We can write 3×3×2 as 3! For short.“3!” Is said “3 factorial”.
FactorialsIn a situation where all of the available
options must be used, but where each can be used only once (i.e. without replacement), the number of possible scenarios is given by:
n! = n × (n-1) × (n-2) × … 3 × 2 × 1
![Page 34: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/34.jpg)
FactorialsIn a situation where all of the available
options must be used, but where each can be used only once (i.e. without replacement), the number of possible scenarios is given by:
n! = n × (n-1) × (n-2) × … 3 × 2 × 1
E.g.: What is the value of: a 5! b 10!
a 5! = 5 4 3 2 1
= 120
b 10! Use a calculator — use the n! key. = 3 628 800
This is said as: “n factorial”
Have a go:Do Sigma (new): Ex. 8.01 (pg. 150)
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Lesson 2: Permutations
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Warm-up quiz(1.) Calculate 9!.
(2.) Calculate 9! 8!
(3.) Write in simplest form: n! (n-1)!
(4.) There are 5 rooms in a house. Each has a single light with a switch. If each light can be either off or on, how many different possibilities are there?
9! = 9×8×7×…×2×1 = 362880 answer
9! = 9×8×7×6×5×4×3×2×1 8! 8×7×6×5×4×3×2×1
Questions from HW?
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Warm-up quiz(1.) Calculate 9!.
(2.) Calculate 9! 8!
(3.) Write in simplest form: n! (n-1)!
(4.) There are 5 rooms in a house. Each has a single light with a switch. If each light can be either off or on, how many different possibilities are there?
9! = 9×8×7×…×2×1 = 362880 answer
9! = 9×8×7×6×5×4×3×2×1 = 9 answer8! 8×7×6×5×4×3×2×1
n! = n×(n-1)×(n-1)×.. ×3×2×1 (n-1)! (n-1)×(n-1)×.. ×3×2×1
Questions from HW?
![Page 38: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/38.jpg)
Warm-up quiz(1.) Calculate 9!.
(2.) Calculate 9! 8!
(3.) Write in simplest form: n! (n-1)!
(4.) There are 5 rooms in a house. Each has a single light with a switch. If each light can be either off or on, how many different possibilities are there?
9! = 9×8×7×…×2×1 = 362880 answer
9! = 9×8×7×6×5×4×3×2×1 = 9 answer8! 8×7×6×5×4×3×2×1
n! = n×(n-1)×(n-1)×.. ×3×2×1 = n answer(n-1)! (n-1)×(n-1)×.. ×3×2×1
Use the Multiplication Principle:Total number poss. outcomes = nbr poss. outcomes for room 1 × nbr poss outcomes
for room 2 etc.
Questions from HW?
![Page 39: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/39.jpg)
Warm-up quiz(1.) Calculate 9!.
(2.) Calculate 9! 8!
(3.) Write in simplest form: n! (n-1)!
(4.) There are 5 rooms in a house. Each has a single light with a switch. If each light can be either off or on, how many different possibilities are there?
9! = 9×8×7×…×2×1 = 362880 answer
9! = 9×8×7×6×5×4×3×2×1 = 9 answer8! 8×7×6×5×4×3×2×1
n! = n×(n-1)×(n-1)×.. ×3×2×1 = n answer(n-1)! (n-1)×(n-1)×.. ×3×2×1
Use the Multiplication Principle:Total number poss. outcomes = 2 × 2 × 2 × 2 × 2
= 25
= 32 possibilities. answer
Questions from HW?
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Sigma Ex. 8.01 - Q14In morse code, messages can be sent by various groups of short and long beeps. If
groups of up to 5 beeps can be transmitted, how many groups are possible?Nbr of possibilities if 1 beep is transmitted, 2 beeps, … etc. up to 5 beeps:1 beep: 2 possibilities (short or long)2 beeps: 2×2 (2 for first beep × 2 for second) = 22
3 beeps: 2 × 2 × 2 = 23
4 beeps: 2 × 2 × 2 × 2 = 24
5 beeps: 2 × 2 × 2 × 2 × 2 = 25
i.e. 2 to the power of the number of beeps in the transmission.
So nbr different arrangements ofbeeps possible for groups of up to 5 = Nbr ways for 1 beep + Nbr ways
for 2 + … Nbr ways for 5 beeps.
= 2 + 22 + 23 + 24 + 25
= 62 answer
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Sigma Ex. 8.01 - Q14In morse code, messages can be sent by various groups of short and long beeps. If
groups of up to 5 beeps can be transmitted, how many groups are possible?Nbr of possibilities if 1 beep is transmitted, 2 beeps, … etc. up to 5 beeps:1 beep: 2 possibilities (short or long)2 beeps: 2×2 (2 for first beep × 2 for second) = 22
3 beeps: 2 × 2 × 2 = 23
4 beeps: 2 × 2 × 2 × 2 = 24
5 beeps: 2 × 2 × 2 × 2 × 2 = 25
i.e. 2 to the power of the number of beeps in the transmission.
So nbr different arrangements ofbeeps possible for groups of up to 5 = Nbr ways for 1 beep + Nbr ways
for 2 + … Nbr ways for 5 beeps.
= 2 + 22 + 23 + 24 + 25
= 62 answer
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Permutations and Combinations
PermutationsPermutations(Arrangements)
• Order matters.
• WHAT you select matters, AND…
• HOW you arrange it:– A,B,C is different to
A,C,B.– E.g. Number plates:
PL9684 & PL4896.
CombinationsCombinations(Selections)
• Order is irrelevant.
• WHAT you select is ALL that matters.
– E.g. selecting 20 leaders from 100 students.
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How to KNOW whether it’s a permutationpermutationor a combinationcombination:
Does ORDER matter?
YESYES – It matters whether – It matters whetherit’s A,B,C, or A,C,Bit’s A,B,C, or A,C,B
NONO – it – it doesn’tdoesn’t matter matterwhether it’s A,B,C or A,C,Bwhether it’s A,B,C or A,C,B
PermutationsPermutations((ArrangingArranging))
CombinationsCombinations((SelectingSelecting only) only)
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Which of these are dealing with Permutations (arrangements) and which with Combinations
(selections)??
1. Mrs O’Brien must choose a College Captain and a Student Rep for the Board from a list of 11 leaders. In how many ways can she do this?
2. A rugby coach must choose a squad of 22 players from a pool of 40 players and assign them positions. In how many ways could he do this?
3. Mrs. O’Brien must select 11 leaders for next year from 50 students returning for Year 13. How many different groups of 11 students is it possible to pick?
4. A cricket coach in a social team must pick a captain and a vice-captain from his 11 players for the following Saturday. In how many ways can he select his captain and his vice-captain?
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Which of these are dealing with Permutations (arrangements) and which with Combinations
(selections)??
5. How many different “words” with 3 letters can be formed from the letters in the word TANK?
6. Graham Henry has 5 contenders for the position of lock in his starting line-up. In how many ways can he select 2 locks from the 5?
7. A cricket coach in a social team must pick a playing 11 for the following Saturday from a squad of 13 players. How many different selections are possible?
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Answers:
1. Permutations.
2. Permutations.
3. Combinations.
4. Permutations.
5. Permutations.
6. Combinations.
7. Combinations.
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How many ways are there of choosing a captain and a vice-captain for a cricket team of eleven players? (Selectors may require this information to write down the different possibilities and then choose the best.)
Using the multiplication principle (like on a tree diagram)
There are two places to fill:
Captain Vice-captain
The captain can be chosen in 11 ways
Once the captain is chosen there are 10 players left to choose the vice-captain from
11 10
11 10Total number of ways: = 110
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Or, we could say there are 110 possible arrangements (Permutations).
Using the Multiplication Principle can get messy with more complex scenarios. There is an easier way to find the number of Permutations or arrangements possible…
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Number of Permutations possible when n options are arranged in
groups of size r:
)!(
!
rn
nPr
n
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In the case of the cricket example:
• The number of Permutations of 11 players, in groups of size 2 (captain and VC) is:
1011123...9
123...91011
)!211(
!112
11
P
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How many ways are there of choosing 3 books from a shelf of 8, and reading them in order?
!
( )!n
rnP n rWe use the formula
n = 8 and r = 3
83P
336
8!5!
40320120
8!
(8 3)!
Multiply out the factorials.
8!5!
8 7 6 5 4 3 2 1
5 4 3 2 1
8 7 6
336
A better way of evaluating 8!5!
Cancel.
Do Sigma (new edition): Pg. 154 - Ex. 8.02(Skip Q13 – or leave it till last)
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Practice examples involving Permutations
Sigma - Page 154
Ex. 8.02
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Extension session
1. Extension to the morse code question from Ex. 8.01.
2. Circular arrangement problems.
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Sigma Ex. 8.01 - Q14In morse code, messages can be sent by various groups of short and long beeps. If
groups of up to 5 beeps can be transmitted, how many groups are possible?Nbr of possibilities if 1 beep is transmitted, 2 beeps, … etc. up to 5 beeps:1 beep: 2 possibilities (short or long)2 beeps: 2×2 (2 for first beep × 2 for second) = 22
3 beeps: 2 × 2 × 2 = 23
4 beeps: 2 × 2 × 2 × 2 = 24
5 beeps: 2 × 2 × 2 × 2 × 2 = 25
i.e. 2 to the power of the number of beeps in the transmission.
So nbr different arrangements ofbeeps possible for groups of up to 5 = Nbr ways for 1 beep + Nbr ways
for 2 + … Nbr ways for 5 beeps.
= 2 + 22 + 23 + 24 + 25
= 62 answer
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Sigma Ex. 8.01 - Q14In morse code, messages can be sent by various groups of short and long beeps. If
groups of up to 5 beeps can be transmitted, how many groups are possible?Nbr of possibilities if 1 beep is transmitted, 2 beeps, … etc. up to 5 beeps:1 beep: 2 possibilities (short or long)2 beeps: 2×2 (2 for first beep × 2 for second) = 22
3 beeps: 2 × 2 × 2 = 23
4 beeps: 2 × 2 × 2 × 2 = 24
5 beeps: 2 × 2 × 2 × 2 × 2 = 25
i.e. 2 to the power of the number of beeps in the transmission.
So nbr different arrangements ofbeeps possible for groups of up to 5 = Nbr ways for 1 beep + Nbr ways
for 2 + … Nbr ways for 5 beeps.
= 2 + 22 + 23 + 24 + 25
= 62 answer
*Extension challenge:1.Write an expression for the number of groups of short or long beeps possible if up to n beeps can be transmitted? 2.What if there are d different types of beeps (i.e. not just short/long)?
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HOW TO SOLVE CIRCULAR ARRANGEMENT PROBLEMS
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8.01H
Draw a diagram to show the number of ways in which 4 people can sit around a round table, in relation to one another. How many different ways are there?
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Draw a diagram to show the number of ways in which King Arthur and 3 of his knights can sit around the round table, in relation to one another. How many different ways are there?
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Sir Lancalot
Sir Galahad
Sir Kay
King Arthur
ROUNDTABLE
Draw a diagram to show the number of ways in which King Arthur and 3 of his knights can sit around the round table, in relation to one another. How many different ways are there?
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What really matters:A circular sequence is not like a linear sequence of ten
positions because it folds back onto itself! Wherever you start from on the circle, you can follow through a
complete sequence!
So all that really matters is the order of seating. We want to know how many different possible orders there are.
E.g. If King Arthur moves one place to the right, but so does everyone else, then the order has not changed.
In short: Only positions with different orders count.
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8.01H
C
A
DB
Since the table is round we need to fix the position of one person as a reference. King Arthur stays put on his thrown!
The following arrangements are all the same, being simple rotations of each other
D
B
AC
C
A
DB
B
D
CA
Draw diagrams showing the different ways with A at the bottom.
Draw a diagram to show the number of ways in which King Arthur and 3 or his knights can sit around the round table, in relation to one another. How many different ways are there?
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Sir Lancalot
Sir Galahad
Sir Kay
King Arthur
ROUNDTABLE
Draw a diagram to show the number of ways in which King Arthur and 3 or his knights can sit around the round table, in relation to one another. How many different ways are there?
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8.01H
A
C
BD
A
D
CB
A
B
DC
A
C
DB
A
D
BC
A
B
CD
and reflection
and reflection
and reflection
D at top B at top
Clearly there are 6 ways.
Draw a diagram to show the number of ways in which King Arthur and 3 of his knights can sit around the round table, in relation to one another. How many different ways are there?
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We regard A as being fixed. King Arthur stays put on his thrown!
A
C
BD
Fixed
Arrange these 3 placesThe other three people can then be consideWhite as being in a line.
A B C D
FixedArrange these 3 places
Total number of ways = (4 – 1)! = 3! = 6
Draw a diagram to show the number of ways in which King Arthur and 3 of his knights can sit around the round table, in relation to one another. How many different ways are there?
King Arthur
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So for any circle problem:
Q: In how many ways can n objects be arranged around a circle?
• Fix one object – it may help to think of King Arthur staying put on his throne.
• In how many ways can the other objects be arranged around it?
• Answer: (n-1)!
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What’s the difference between a circular arrangement 4 objects (A–D) and a linear one?
NORMAL LINEAR SEQUENCE
(Has a beginning & end)
These represent 4 Different Arrangements
A, B, C, D
D, A, B, C
C, D, A, B
B, C, D, A
CIRCULAR EQUIVALENT(Continuous sequence)
1 Arrangement – it’s just in 4 different positions.
A, B, C, D, A, B, C, …
D, A, B, C, D, A, B, …
C, D, A, B, C, D, A, …
B, C, D, A, B, C, D, …
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There are 4! possible linear arrangements of 4 objects.
So the number of circular arrangements possible for 4 objects is:
44!
44 123
!( )14 123
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So, the number of possible arrangements of n objects around a circle is given by:
!)1(!
nn
nNumber of circular arrangements =
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Lesson 3: Combinations
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How to KNOW whether it’s a permutationpermutationor a combinationcombination:
Does it matter what ORDER the items are ARRANGED into?
YESYES – It matters whether – It matters whetherit’s A,B,C, or A,C,Bit’s A,B,C, or A,C,B
NONO – it – it doesn’tdoesn’t matter matterwhether it’s A,B,C or A,C,Bwhether it’s A,B,C or A,C,B
PermutationsPermutations(Arranging)(Arranging)
CombinationsCombinations(Selecting only)(Selecting only)
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Number of Combinations (possible selections) of r items from
n items:
)!(!
!
rnr
nCr
n
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E.g 1: a Calculate the value of 11C3. b Give an example of a situation involving choice where the number of possible choices would be 11C3.
113C
16511C3 could be the number of different choices of 3 chocolates from a box of 11
This can either be written in fulla
11!
3!(11 3)!
11!3! 8!
11 10 9 8 7 6 5 4 3 2 1(3 2 1) (8 7 6 5 4 3 2 1)
11 10 93 2 1
Cancel.
b
Do Sigma (new edition): Pg. 159 - Ex. 8.03(Skip Q15 – or leave it till last)
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Practice examples involving Combinations
Sigma - Page 159
Ex. 8.03
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1. Mrs O’Brien must select 11 leaders for next year from 42 students returning for Year 13. How many different groups of 11 students is it possible to pick?
She’s selecting 11 items from 42, so the answer is:
Nbr poss selections = 42C11
= 4 280 561 376possible selections.
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Lesson 4: Probabilities involving permutations & combos 1
Warm-up quiz.Work:Do an example on board, then Ex. 8.04 (new
Sigma) – do Q1 & 2 as a class.
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E.g: Your friend lets you choose 3 different colours of lollie from a box containing 5 different colours: White, Green, Blue, Yellow and Orange.1.How many different selections of 3 are possible?2.What is the probability that a random selection of 3 of these 5 lollies includes the White one?
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E.g: Your friend lets you choose 3 different colours of lollie from a box containing 5 different colours: White, Green, Blue, Yellow and Orange.1.How many different selections of 3 are possible?
!
!( )!n
rnC r n r
n = 5 r = 3
35C
)!35(!3
!5
!2!3
!5
12
120
10 possible selections of 3 lollies from 5.
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possible selections of 3 lollies from 5.
E.g: Your friend lets you choose 3 different colours of lollie from a box containing 5 different colours: White, Green, Blue, Yellow and Orange.1.How many different selections of 3 are possible?
35C 10
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possible selections of 3 lollies from 5.
E.g: Your friend lets you choose 3 different colours of lollie from a box containing 5 different colours: White, Green, Blue, Yellow and Orange.1.How many different selections of 3 are possible?2.What is the probability that a random selection of 3 of these 5 lollies includes the White one?
35C 10
Calculating the probability of getting a particular selection or arrangement of items:
Formula:
Total number of possible outcomes
P(Desired outcome) = Number of ways of getting
desired outcome
What if order matters?
Do Sigma (new edition): Ex. 8.04
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Ex. 8.04
Page 161
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
? ? ? ?
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
? ? ? ?
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 ? ? ?
9 possibilities for the first digit
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 ? ? ?
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 8 ? ?
8 possibilities for the second digit – one used up
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 8 ? ?
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 8 7 ?
7 possibilities for the third digit – 2 used up
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 8 7 ?
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
9 8 7 6
6 possibilities for the third digit – 3 used up
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Sigma: Page 161Ex 8.04
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
(a)How many different PIN numbers are possible?
Without replacement and order matters - arranging (not just selecting). permutations.
)!(
!n
rn
nPr
= 3024 different PINS are possible.
!5
!94
9 P Same as saying: 9 × 8 × 7 × 6
9 8 7 6
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Sigma: Page 161Ex 8.04
? ? ? ?
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5
Sigma: Page 161Ex 8.04
? ? ? ?
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Sigma: Page 161Ex 8.04
5 ? ? ?
![Page 94: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/94.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Sigma: Page 161Ex 8.04
5 ? ? ?
![Page 95: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/95.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill.
Sigma: Page 161Ex 8.04
5 ? ? ?
![Page 96: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/96.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 ? ? ?
![Page 97: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/97.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 ? ? ?
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 8 ? ?
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 8 ? ?
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 8 7 ?
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 8 7 ?
![Page 102: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/102.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.
Sigma: Page 161Ex 8.04
5 8 7 6
![Page 103: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/103.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations.
Sigma: Page 161Ex 8.04
5 8 7 6
![Page 104: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/104.jpg)
1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3
Nbr poss. PINS starting with an odd digit = 5 × 8P3
Sigma: Page 161Ex 8.04
5 8 7 6
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3
Nbr poss. PINS starting with an odd digit = 5 × 8P3
or 5 × 8 × 7 × 6
Sigma: Page 161Ex 8.04
5 8 7 6
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(b)How many of the PIN numbers start with an odd digit?
Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)
Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3
Nbr poss. PINS starting with an odd digit = 5 × 8P3
or 5 × 8 × 7 × 6 = 1680 PINS.
Sigma: Page 161Ex 8.04
5 8 7 6
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1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.
…(c) Calculate the probability that a client is given a PIN number
starting with an odd digit
Nbr poss. PINS starting with an odd digit (from b) = 5 × 8P3
= 1680
Sigma: Page 161 - Ex 8.04
Total number poss. PIN numbers (from a) = 9P4
= 3024
P(1st digit is odd) = numbers PIN poss.number Total
digit oddan with starting PINS poss.Nbr
3024
1680 =
9
5 = answer
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Lesson 5: Probabilities involving permutations & combos 2
Work:Go over some Qs from Ex. 8.04.Continue onto Ex. 8.05.
HW: Finish to the end of Ex. 8.05.
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The birthday problem1.) In a room of 5 people, what is the probability that at least
one pair shares the same birthday?
2.) And in a room of 30 people?
3.) Write an expression for the probability that at least one pair from n people shares the same birthday?
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6. (a) Complete the probability distribution table for the nbr of White balls obtained when 3 balls are selected at random from an urn containing 6 White and 4 yellow balls. (NCEA Merit level)
P(X=0)
Sigma: Page 161 - Ex 8.04
P(selecting 0 White balls) = balls 3 of selections poss. ofnumber Total
yellow 3 selecting of Ways red 0 selecting of Ways
310
34
06
C
CC =
=
Nbr ways of selecting 3 of the 4 yellow balls = 4C3 - i.e. 4.
30
1
Nbr ways of selecting 0 of the 6 White balls = 6C0 - i.e. 1.
Let X: number of White balls selected.
x 0 1 2 3
P(X=x)30
1
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6. (a) Complete the probability distribution table for the nbr of White balls obtained when 3 balls are selected at random from an urn containing 6 White and 4 yellow balls. (NCEA Merit level)
P(X=1)
Sigma: Page 161 - Ex 8.04
P(selecting 1 White ball) = balls 3 of selections poss. ofnumber Total
yellow 2 selecting of Ways red 1 selecting of Ways
310
24
16
C
CC =
=
Nbr ways of selecting 2 of the 4 yellow balls = 4C2 - i.e. 6.
10
3
Nbr ways of selecting 1 of the 6 White balls = 6C1 - i.e. 6.
Let X: number of White balls selected.
x 0 1 2 3
P(X=x)30
1
30
9
30
9
or
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x 0 1 2 3
P(X=x)
6. (a) Complete the probability distribution table for the nbr of White balls obtained when 3 balls are selected at random from an urn containing 6 White and 4 yellow balls. (NCEA Merit level)
P(X=2)
Sigma: Page 161 - Ex 8.04
P(selecting 2 White ball) = balls 3 of selections poss. ofnumber Total
yellow 1 selecting of Ways red 2 selecting of Ways
310
14
26
C
CC =
=
Nbr ways of selecting 1 of the 4 yellow balls = 4C1 - i.e. 4.
2
1
Nbr ways of selecting 2 of the 6 White balls = 6C2 - i.e. 15.
Let X: number of White balls selected.
30
1
30
15
30
9
or
30
15
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x 0 1 2 3
P(X=x)
6. (a) Complete the probability distribution table for the nbr of White balls obtained when 3 balls are selected at random from an urn containing 6 White and 4 yellow balls. (NCEA Merit level)
P(X=3)
Sigma: Page 161 - Ex 8.04
P(selecting 3 White balls) = balls 3 of selections poss. ofnumber Total
yellow 0 selecting of Ways red 3 selecting of Ways
310
04
36
C
CC =
=
Nbr ways of selecting 0 of the 4 yellow balls = 4C0 - i.e. 1.
6
1
Nbr ways of selecting 3 of the 6 White balls = 6C3 - i.e. 20.
Let X: number of White balls selected.
30
1
30
5
30
9
or
30
15
30
5
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x 0 1 2 3
P(X=x)
6. (b) If X is the number of White balls obtained, find an expression for P(X=x). The expression should use combinations.
(NCEA Excellence level)
P(X= x)
Sigma: Page 161 - Ex 8.04
P(selecting x White balls) = balls 3 of selections poss. ofnumber Total
yellow x)-(3 selecting of Ways red x selecting of Ways
310
)3(46
C
CC xx = answer
Nbr ways of selecting (3- x) of the 4 yellow balls = 4C(3-x)
Nbr ways of selecting x of the 6 White balls = 6Cx
Let X: number of White balls selected.
30
130
9
30
15
30
5
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Lesson 6: Revision of all covered so far in probability
STARTER: The birthday problem
Go over some Qs from Ex. 8.04 and 8.05.
TEST TOMORROW ON EVERYTHING DONE SO FAR IN PROBABILITY:
Revision: Do Sigma (2nd edition): Ex. 1.6 (p19)Or Ex. 9.05 (p178) in the new edition of Sigma.
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Extension lesson – Combinations and Pascal’s Triangle
1. Go over how the Combinations formula is derived from the Permutations formula.
2. Intro to Pascal’s Triangle: Check out http://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif (Moving demo of how the triangle is formed).
3. Draw Pascal’s Triangle.4. How to use it to find nCr. Application: NuLake Q132
HW: Prove that each number in Pascal’s is formed by adding the 2 numbers above it: i.e. nCr-1 + nCr = n+1Cr
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What’s really going on with combinations?
The next 4 slides are extra for experts.
They explain what is really going on with combinations – i.e. how we get the formula:
for the number of possible selections of r objects from n objects.
)!(!
!
rnr
nCr
n
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EXPLANATION FOR COMBINATION FORMULA (optional extra):
Combinations (selections) do not take order into account. Permutations do.
So number of combinations will be the number of permutations divided by the variations in order.
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Take the cricket example: What if we must simply select two players
from a team of 11. Positions (and thus order) are irrelevant so we’re
dealing with Combinations.
So there are 11C2 possible combinations.
12.. 2
11
211
PCei
ncombinatioper possible ordersdifferent ofNumber
nspermutatio ofNumber nscombinatioNumber
!2211 P
![Page 120: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/120.jpg)
Take the cricket example: What if we must simply select two players
from a team of 11. Positions (and thus order) are irrelevant so we’re
dealing with Combinations.
So there are 11C2 possible combinations.
12.. 2
11
211
PCei
ncombinatioper possible ordersdifferent ofNumber
nspermutatio ofNumber nscombinatioNumber
!2)!211(
!11
!2211
P
![Page 121: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/121.jpg)
Take the cricket example: What if we must simply select two players
from a team of 11. Positions (and thus order) are irrelevant so we’re
dealing with Combinations.
So there are 11C2 possible combinations.
12.. 2
11
211
PCei
ncombinatioper possible ordersdifferent ofNumber
nspermutatio ofNumber nscombinatioNumber
!2
1
)!211(
!11
!2211
P
![Page 122: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/122.jpg)
Take the cricket example: What if we must simply select two players
from a team of 11. Positions (and thus order) are irrelevant so we’re
dealing with Combinations.
So there are 11C2 possible combinations.
12.. 2
11
211
PCei
ncombinatioper possible ordersdifferent ofNumber
nspermutatio ofNumber nscombinatioNumber
)!211(!2
!11
!2
1
)!211(
!11
!2211
P
![Page 123: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/123.jpg)
12.. 2
11
211
PCei
)!211(!2
!11
!2
1
)!211(
!11
!2211
P
And in general, when selecting r items from n items…
rnC
rnP !rSo we get
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!2.. 211
211 PCei
And in general, when selecting r items from n items…
rnC
rnP !rSo we get
)!211(!2
!11
!2
1
)!211(
!11
!r
)!(
!
rn
n
![Page 125: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/125.jpg)
rnC
ncombinatio per possible orders different of Number nspermutatio of Number
nscombinatio Number
And in general, when selecting r items from n items…
)!(
!
rn
n
)!(!
!
rnr
n
r
nP !r
!
1
r
So we get
![Page 126: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/126.jpg)
Pascal’s Triangle
![Page 127: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/127.jpg)
Pascal’s Triangle
Intro to Pascal’s Triangle: Check out http://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif
![Page 128: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/128.jpg)
Pascal’s Triangle
![Page 129: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/129.jpg)
Pascal’s Triangle
![Page 130: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/130.jpg)
Extension challenge:Prove using factorials that each number in Pascal’s Triangle is the sum of the 2 adjacent numbers in the row above:
That is, prove thatr
nr
nr
n CCC 11
![Page 131: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/131.jpg)
Prove using factorials that:
LHS rn
rn CC 1
rn
rn
rn CCC 1
1
)!(!
!
))!1(()!1(
!
rnr
n
rnr
n
)!(!
!
)!1()!1(
!
rnr
n
rnr
n
)1()!(!
)1(!
)!1()!1(
!
rnrnr
rnn
rnrr
rn
)!1(!
)1(!
)!1(!
!
rnr
rnn
rnr
rn
)!1(!
)1(!
rnr
rnrn
)!1(!
)!1(
rnr
n
rn C1 = RHS as required.
× r
× r
× (n+1-r)
× (n+1-r)
![Page 132: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/132.jpg)
Five baseball players threw their caps into a sport’s bag after a ball game. What is the probability that each person will pick his or her own hat when starting the next game?
![Page 133: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/133.jpg)
The probability that the first person will get the right hat is 1/5. Now there are four people and four hats so the probability that the second person gets their own hat is…?
HINT
1/4
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The total probability of everyone getting his/her own hat is found by multiplying all the possibilities together.(1/5) x (1/4) x (1/3) x (1/2) x1 =
1/(5x4x3x2x1) =
![Page 135: Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.](https://reader033.fdocuments.net/reader033/viewer/2022051621/56649ec95503460f94bd62d5/html5/thumbnails/135.jpg)
So the probability that each person will pick his or her own hat is...
1/120
End of Activity