Consider the word:

RANDOM

How many actual words longer than two letters can you find from the English Language using the letters in the word RANDOM?

Lesson Objective

Be able to find the number of ways of arranging a number of items in a list. Use factorial notation to describe these situations .

53 words Found

6 Letter Scrabble Words

random rodman

5 Letter Scrabble Words

adorn andro manor monad nomad radon roman

4 Letter Scrabble Words

damn darn dona dorm dram mano moan mora morn

nard noma norm orad rand road roam roan

3 Letter Scrabble Words

ado and arm dam dan dom don dor mad man mar

moa mod mon mor nam nod nom nor oar oda ora rad

ram ran rod rom

http://www.scrabblefinder.com/solver/

Suppose we didn’t need to make proper words - just arrangements of letters (like a secret code) – how many different ways could we

arrange the letters in the word RANDOM?

Suppose we didn’t need to make proper words - just arrangements of letters (like a secret code) – how many different ways could we

arrange the letters in the word RANDOM?

Start with 2 letters

RA AR

Suppose we didn’t need to make proper words - just arrangements of letters (like a secret code) – how many different ways could we

arrange the letters in the word RANDOM?

Start with 2 letters

RA AR

3 letters

RAN ARN NAR

RNA ANR NRA

Suppose we didn’t need to make proper words - just arrangements of letters (like a secret code) – how many different ways could we

arrange the letters in the word RANDOM?

Start with 2 letters

RA AR

3 letters

RAN ARN NAR

RNA ANR NRA

4 letters

RAND RADN RDAN ANRD ARND ADNR

RDNA RNDA RNAD ANDR ARDN ADRN

NDRA NRAD NARD DANR DNAR DRNA

NDAR NRDA NADR DARN DNRA DRAN

How many ways can you arrange the letters in the word: PUPIL

How many ways can you arrange the letters in the word: POPPY?

How many ways can you arrange the letters in the word: BANANA?

How many ways can you arrange the letters in the word: STATISTICS?

How many ways can you arrange the letters in the word: floccinaucinihilipilification ?

Suppose we take the word RANDOM and select 4 different letters to create a password, how many potential passwords can we create?

What if the passwords have to begin with a vowel?

What if the letters in the word can be used multiple times?

1) How many ways can you arrange 8 people in a line at a bus stop?

2) A cricket team has 11 batsmen. If you randomly place them in the batting order how many different batting line ups are possible?

3) A key pad has the 10 digits 0 to 9 on it. a) How many 4 digit codes are possible if you don’t repeat any digits? b) How many 4 digit codes are possible if repeated digits are allowed? c) How many 4 digit codes are possible if you zeros at the start (like 0023) are not

allowed?

4) A shelf has space for 5 books, but I have 8. How many different ways can I stack the books on the shelf?

5) In a race there are 10 competitors.a) How many different ways can the three podium places be filled?

b) If the same race is run with exactly the same competitors the next day, how many ways can the winning places be filled over the two nights?

6) Suppose you have the digits 9, 8, 7, 4, 2. How many 5 digit numbers can you make if no repeated digits are allowed How many 5 digit even numbers can you make if no repeated digits are allowed How many 5 digit even numbers bigger than 50000 can you make (no repeats!)

7) a) How many ways can you sit 6 people around a round table? b) How many ways can you arrange 6 different coloured beds on a bracelet?

Did you know?

that if you spend a few minutes truly shuffling a pack of cards your cards will almost certainly be ordered in a way that is entirely

unique and that has never been obtained before

Did you know?

that if you spend a few minutes truly shuffling a pack of cards your cards will, well beyond reasonable doubt, be ordered in a way that is

entirely unique and that has never been obtained before

There are 52! ways of arranging the cards.This is more than 80 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 ways.

There are only 6 000 000 000 people on the planet if they each shuffled a pack every second of every day for a whole year it would take them 4.2627 x 1050 before a single repeat ordering was expected. Since we have only been shuffling cards for around 2000-3000 years at a much slower rate the probability of a repeat so far is insignificantly small!

The table shows the number of child visitors at an event.What is the mean number of children per family and the standard deviation in the number of children per family?

Number of children in the family

Frequency

1 3

2 4

3 8

4 2

5 3

Without any extra calculation write down how you can find the mean and sd for this set of data:

x Frequency

3 3

5 4

7 8

9 2

11 3

1

22

n

xnx

1

)( 2

n

xx

1 3 3 32 4 8 163 8 24 724 2 8 325 3 15 75

20 58 198

29.81.568421053sd = 1.252366182

Lesson Objective

Understand the difference between an arrangement and a selection

In a classroom there are 8 pupils.

a) How many different ways could I arrange 3 of these pupils in a line?

b) How many different ways could I select 3 pupils out of the eight to be in a team?

What is the difference between these two questions?

Are there more ways to select or to arrange. Why?

1) A swimming team of five is to be selected from a squad of 7. How many possible teams are there?

2) An exam board randomly selects 6 coursework tasks to moderate from a sample of 20. How many different samples could be selected?

3) How many different ways can I select 4 people in this room to take on a trip?

In general the number of ways of selecting ‘r’ things from ‘n’

is - we call this

This represents the number of ways of choosing ‘r’ items from ‘n’.

!)!(

!

rrn

n

r

nC

1) A swimming team of five is to be selected from a squad of 7. How many possible teams are there?

2) An exam board randomly selects 6 coursework tasks to moderate from a sample of 20. How many different samples could be selected?

3) How many different ways can I select 4 people in this room to take on a trip?

1) There are five maths teachers and six english teachers in a meeting.

How many ways are there of choosing a subcommittee of two maths teachers and three english teachers?

This time the subcommittee of five is chosen by drawing names from a hat. What is the probability there are no maths teachers on the subcommittee?

More difficult problems involving selections

(i) Calculate the total number of ways they can be chosen.

(ii) What is the probability that the selected group of has more boys than girls?

2) Four representatives are chosen from a teaching group consisting of 12 boys and 8 girls.

Puzzle 1

There are 7 staff and 6 students on the sports council of a college. A committee of 8 people from the 13 on the council is to be selected to organise a tennis competition. How many different committees of 8 can be selected? What is the probability that the committee selected has more staff than students?

I have a box of chocolates with 10 different chocolates left in it. Of these, there are 6 which I particularly like. However, I intend to offer my three friends one chocolate each before I eat the rest.

How many different selections of chocolates can I be left with after my friends have chosen?

Show that 36 of these selections leave me with exactly 5 chocolates which I particularly like.

How many selections leave me with:

(i) all 6 of the chocolates that I particularly like?

(ii) exactly 4 of the chocolates that I particularly like?

(iii) exactly 3 of the chocolates that I particularly like?

Assuming my friends choose at random, what is the most likely outcome, and what is the probability of that outcome?

Puzzle 2

The formula for is

Find:

What do you notice about these numbers? Why does this happen? Will this always happen?

What do these numbers represent?

What does it tell you about how we define 0! ?

!)!(

!

rrn

n

rnC

07C 1

7C 37C 4

7C 57C2

7C 67C 7

7C

Puzzle 3

Puzzle 4

a) In the National Lottery how many different ways can the 6 balls be selected in the main draw.

b) If you buy one ticket what is the probability that it is a winning ticket?

c) What is the probability that you manage to match exactly 4 out of the 6 balls.

Puzzle 5

At a small bank the manager has a staff of 12, consisting of 5 men and 7 women including a Mr Brown and a Mrs Green. The manager receives a letter from head office saying that 4 of his staff are to be made redundant. In the interests of fairness the manager selects the 4 staff randomly.

a) How many different selections are possible?

b) How many of these will include both Mr Brown and Mrs Green?

c) What is the probability that both Mr Brown and Mrs Green are made redundant?

At the last minute head office decides that equal numbers of men and women should be made redundant. How many selections are there now and what is the probability that Mr Brown and Mrs Green are both made redundant

Puzzle 6

Is it true that ?rnC

111

rn

rn CC =

Can you prove this!

Puzzle 113C8 = 1287 Can have 5 staff 3 stu 7C5 x 6C3 = 420 or 6 2 stu 7C6 x 6C2 = 105 or 7 1 stu 7C7 x 6C1 = 6

So Prob = 531/1287

Puzzle 210C7 = 1287 (how many ways to select 7 from 10) 5 chocs I like = 6C5 x 4C2 (5 from 6 ad 2 others)6 will be 6C6 x 4C1 = 44 will be 6C4 x 4C3 = 603 will be 6C3 x 4C4 = 20So 4 is most likely outcome

Puzzle 31 7 21 35 21 7 1Symmetry – choosing 5 from 7 is like choosing 2 from 7Choosing 0 from 7 can only happen in 1 way choosing 76 from 7 can only happen in 1 way so 0! Must be 1

Puzzle 449C6 = 13983816P(win) = 1/13983816

P(match 4) = 6C4 x 43C2 (ways of getting 4) divided by 13983816= 0.000969 so a little under a 0.1% chance

Puzzle 512C4 = 495Choose Brown Green and any 2 from remaining 10 = 1 x 1 x 10C2 = 45P(Both B and G redundant) = 45/495

5C2 x 7C2 = 210Choose Brown, and extra male, Green and extra female = 1 x 4C1 x 1 x 6C1 = 24

Puzzle 6

See later

++

What is this? What’s it got to do with Combinations and ? r

nC

Note, Pascal’s triangle shows us that:

rnC

111

rn

rn CC =

You can prove this algebraically!

Note, Pascal’s triangle shows us that:

rnC

111

rn

rn CC =

You can prove this algebraically!

You can also prove it using logic:Suppose you need to select ‘r’ items from ‘n’ things, then you can either select the first item and then ‘r-1’ items from the remaining ‘n-1’ things (ie ) or you can not select the first item which means you must select all ‘r’ items from the remaining ‘n-1’ things (ie ).

11

rn C

rn C1