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 .
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?
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
nard
dorm
noma norm
dram mano moan mora
morn
orad rand road roam roan
3 Letter Scrabble Words
ado and arm dam
moa mod
dan
dom
don
dor mad man mar
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
 (x  x)
n 1
x
2
 nx 2
n 1
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
2
1
2
3
4
5
3
4
8
2
3
20
3
8
24
8
15
58
3
16
72
32
75
198
29.8
1.568421053
sd = 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?
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’
n
n
!
is
- we call this C r
(n  r )!r!
This represents the number of ways of choosing ‘r’ items from ‘n’.
More difficult problems involving selections
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?
2) Four representatives are chosen from a teaching group
consisting of 12 boys and 8 girls.
(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?
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?
Puzzle 2
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 3
n!
(n  r )!r!
n
The formula for C r is
7
7
7
C
C1 C2
Find:
0
7
C3 7 C 4
7
C5
7
C6
7
C7
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! ?
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
n
Cr =
Can you prove this!
n 1
Cr 
n 1
Cr 1
?
Puzzle 1
13C8 = 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 4
49C6 = 13983816
P(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 5
12C4 = 495
Puzzle 2
10C7 = 1287 (how many ways to select 7 from 10) Choose Brown Green and any 2 from
remaining 10 = 1 x 1 x 10C2 = 45
5 chocs I like = 6C5 x 4C2 (5 from 6 ad 2 others)
P(Both B and G redundant) = 45/495
6 will be 6C6 x 4C1 = 4
4 will be 6C4 x 4C3 = 60
3 will be 6C3 x 4C4 = 20
So 4 is most likely outcome
5C2 x 7C2 = 210
Choose Brown, and extra male, Green and
extra female = 1 x 4C1 x 1 x 6C1 = 24
Puzzle 3
1 7 21 35 21 7 1
Symmetry – choosing 5 from 7 is like choosing 2 from 7
Choosing 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 6
See later
+
What is this? What’s it got to do with Combinations and
n
Cr ?
Note, Pascal’s triangle shows us that:
n
Cr =
n 1
Cr 
You can prove this algebraically!
n 1
Cr 1
Note, Pascal’s triangle shows us that:
n
Cr =
n 1
Cr 
n 1
Cr 1
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 n1Cr 1 ) or you can not select the first item which
means you must select all ‘r’ items from the remaining ‘n-1’
things (ie n1Cr ).