Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.

Statistics and Modelling Course

ARRANGEMENTS&

SELECTIONS

Part of Achievement Standard 90643 - Probability

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.

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.

If they go to Sumner:

Fast food restaurant

Movie complex

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.

Beach Fast food restaurant Movie complex

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.

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

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

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.

The possibilities are shown in the diagram.

Choose A as the 1st person.

Once the 1st person has been

chosen, there are two ways of

choosing the 2nd person.

Once the 2nd person has been

chosen, there is only one way of choosing the 3rd person.

Choose B as the 1st person.

Choose C as the 1st person.

From the diagram, there are 6 ways altogether

Using the Multiplication Principle There are three positions to fill:

Using the Multiplication Principle

The first position can be filled in 3 ways

There are three positions to fill:

Once the first person is chosen, the second person can be chosen in any one of 2 ways as one person is already used

8.01E Alex, Barbara and Colin are waiting to enter a cave. In how many ways can they enter the cave in single file?

Once the second person is chosen, the third person can be chosen in only one way

Number of possible orders =

We can write 3×2×1 as 3! For short.“3!” Is said “3 factorial”.

Factorials

Using the multiplication principle:

3 2 1 Number of possible orders =

Factorials

FactorialsIn a situation where all of the available

options must be used,

options must be used, but where each can be used only once (i.e. without replacement),

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

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

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)

Lesson 2: Permutations

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?

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?

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?

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

= 32 possibilities. answer

Questions from HW?

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

3 beeps: 2 × 2 × 2 = 23

4 beeps: 2 × 2 × 2 × 2 = 24

5 beeps: 2 × 2 × 2 × 2 × 2 = 25

= 2 + 22 + 23 + 24 + 25

= 62 answer

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.

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)

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?

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?

Answers:

1. Permutations.

2. Permutations.

3. Combinations.

4. Permutations.

5. Permutations.

6. Combinations.

7. Combinations.

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 10Total number of ways: = 110

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…

Number of Permutations possible when n options are arranged in

groups of size r:

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(

How many ways are there of choosing 3 books from a shelf of 8, and reading them in order?

rnP n rWe use the formula

n = 8 and r = 3

40320120

(8 3)!

Multiply out the factorials.

8 7 6 5 4 3 2 1

5 4 3 2 1

A better way of evaluating 8!5!

Cancel.

Do Sigma (new edition): Pg. 154 - Ex. 8.02(Skip Q13 – or leave it till last)

Practice examples involving Permutations

Sigma - Page 154

Ex. 8.02

Extension session

1. Extension to the morse code question from Ex. 8.01.

2. Circular arrangement problems.

3 beeps: 2 × 2 × 2 = 23

4 beeps: 2 × 2 × 2 × 2 = 24

5 beeps: 2 × 2 × 2 × 2 × 2 = 25

= 2 + 22 + 23 + 24 + 25

= 62 answer

3 beeps: 2 × 2 × 2 = 23

4 beeps: 2 × 2 × 2 × 2 = 24

5 beeps: 2 × 2 × 2 × 2 × 2 = 25

= 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)?

HOW TO SOLVE CIRCULAR ARRANGEMENT PROBLEMS

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?

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?

Sir Lancalot

Sir Galahad

Sir Kay

King Arthur

ROUNDTABLE

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.

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

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?

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?

and reflection

D at top B at top

Clearly there are 6 ways.

We regard A as being fixed. King Arthur stays put on his thrown!

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

King Arthur

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)!

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, …

There are 4! possible linear arrangements of 4 objects.

So the number of circular arrangements possible for 4 objects is:

44 123

!( )14 123

So, the number of possible arrangements of n objects around a circle is given by:

nNumber of circular arrangements =

Lesson 3: Combinations

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)

Number of Combinations (possible selections) of r items from

n items:

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.

16511C3 could be the number of different choices of 3 chocolates from a box of 11

This can either be written in fulla

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.

Do Sigma (new edition): Pg. 159 - Ex. 8.03(Skip Q15 – or leave it till last)

Practice examples involving Combinations

Sigma - Page 159

Ex. 8.03

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.

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.

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?

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

)!35(!3

10 possible selections of 3 lollies from 5.

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

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

Ex. 8.04

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

9 ? ? ?

9 8 ? ?

8 possibilities for the second digit – one used up

9 8 ? ?

9 8 7 ?

7 possibilities for the third digit – 2 used up

9 8 7 ?

9 8 7 6

6 possibilities for the third digit – 3 used up

= 3024 different PINS are possible.

9 P Same as saying: 9 × 8 × 7 × 6

9 8 7 6

…(b)How many of the PIN numbers start with an odd digit?

? ? ? ?

Number of possibilities for first digit = 5

? ? ? ?

Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)

5 ? ? ?

Then there are 3 remaining positions to fill.

5 ? ? ?

Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.

5 ? ? ?

5 8 ? ?

5 8 7 ?

5 8 7 6

Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations.

5 8 7 6

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

5 8 7 6

or 5 × 8 × 7 × 6

5 8 7 6

or 5 × 8 × 7 × 6 = 1680 PINS.

5 8 7 6

…(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

1680 =

5 = answer

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.

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?

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)

P(selecting 0 White balls) = balls 3 of selections poss. ofnumber Total

yellow 3 selecting of Ways red 0 selecting of Ways

Nbr ways of selecting 3 of the 4 yellow balls = 4C3 - i.e. 4.

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

P(X=1)

P(selecting 1 White ball) = balls 3 of selections poss. ofnumber Total

x 0 1 2 3

P(X=x)30

x 0 1 2 3

P(X=x)

P(X=2)

P(selecting 2 White ball) = balls 3 of selections poss. ofnumber Total

x 0 1 2 3

P(X=x)

P(X=3)

P(selecting 3 White balls) = balls 3 of selections poss. ofnumber Total

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)

P(selecting x White balls) = balls 3 of selections poss. ofnumber Total

yellow x)-(3 selecting of Ways red x selecting of Ways

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

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.

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

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.

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.

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

ncombinatioper possible ordersdifferent ofNumber

nspermutatio ofNumber nscombinatioNumber

!2211 P

12.. 2

!2)!211(

12.. 2

)!211(

12.. 2

)!211(!2

)!211(

12.. 2

)!211(!2

)!211(

And in general, when selecting r items from n items…

rnP !rSo we get

!2.. 211

211 PCei

rnP !rSo we get

)!211(!2

)!211(

ncombinatio per possible orders different of Number nspermutatio of Number

nscombinatio Number

So we get

Pascal’s Triangle

Intro to Pascal’s Triangle: Check out http://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif

Pascal’s Triangle

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

n CCC 11

Prove using factorials that:

LHS rn

rn CC 1

rn CCC 1

))!1(()!1(

)!1()!1(

)1()!(!

)!1()!1(

rn C1 = RHS as required.

× (n+1-r)

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?

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…?

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) =

So the probability that each person will pick his or her own hat is...

End of Activity

Statistics and Modelling Course 2011 ARRANGEMENTS & SELECTIONS Part of Achievement Standard 90643 - Probability.

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