Transcript KS3: Straight Lines

Year 8: Combinatorics
Dr J Frost ([email protected])
www.drfrostmaths.com
Last modified: 29th December 2015
PART 1 :: Systematic Counting
Sometimes the easiest way to count things is to explicitly list out every possibility!
The key is listing the possibilities in some systematic way that avoids missing out cases
or duplicates.
Starter
[JMC 2005 Q17] The figure shows rectangle
𝑃𝑅𝑆𝑈 and line 𝑄𝑇, which divides the rectangle
into two squares. How many right-angled
triangles can be drawn using any three of the
points 𝑃, 𝑄, 𝑅, 𝑆, 𝑇, 𝑈 as corners?
Hint: Perhaps consider each possible shape of
triangle and how many there are of each?
×2
×8
×
?4
14 possibilities. By breaking the problem down into types of triangle, it
made it much easier for us to count without missing out any
possibilities. It also prevented us having to draw out similar cases.
Further Example
A frog has 6 lily pads in front of him in a line. He can either hop (H) to the next lily
pad, or skip one and go to the next (S).
List out (using sequences of S and H) all the ways for the frog to get to the final
lily pad. e.g. HSSH.
Easiest when we break down into different numbers of skips:
0 skips: HHHHHH (1 way)
1 skip: SHHHH, HSHHH, HHSHH, HHHSH, HHHHS (5 ways)
2 skips: SSHH, SHSH, SHHS, HSSH, HSHS, HHSS (6 ways)
To ensure I got all the possibilities here, I gradually moved
3 skips: SSS (1 way)
13 ways.
?
the second S right one, before moving the left S right and
repeating:
SSHH
HSSH
HHSS
SHSH
HSHS
SHHS
In summary, break the problem down into smaller problems where it’s easy to
count within each one, e.g. types of triangles, team sizes, number of skips, etc.
Exercise 1
1 At a restaurant, there is a choice of Avocado
(A), Beans (B) or Cauliflower (C) for starter, and
a choice of Dog (D), Escalopes (E) and Frog
Legs (F) for main course. List out all the nine
possible combinations of starter and main
course (e.g. ‘BF’).
? BE, BF, CD, CE, CF.
Solution: AD, AE, AF, BD,
2 [JMC 2006 Q4] How many triangles of any size
are there in this diagram?
A 8
B 10
C 12
D 14
E 16
Solution:
? C
3 [JMC 2013 Q12] How many hexagons are there
in the diagram?
A 4
B 6
C 8
D 10
E 12
Solution:
?E
Exercises on provided sheet.
4 [SMC 2001 Q3] The diagram shows
a regular hexagon divided up into
six equilateral triangles. How many
quadrilaterals are there in the
diagram?
A 6
B 8
C 10
D 12
E 14
Solution: D
?
5 [JMC 2006 Q17] In how many
different ways can a row of five
“on/off” switches be set so that no
two adjacent switches are in the
“off” position?
A 5
B 10
C 11
D 13
E 15
Solution: D
?
Exercise 1
6
[Kangaroo Pink 2005 Q8] In the diagram
there are 7 squares. What is the difference
between the number of triangles and the
number of squares in the diagram?
A 0
B 1
C 2
D 3
E 4
Solution: D
Exercises on provided sheet.
9 [TMC Final 2012 Q6] Find the
number of squares formed by the
lines of this 5 by 7 rectangular grid of
squares.
?
7
[JMO 2011 A6] The diagram shows a grid
of 16 identical equilateral triangles. How
many different rhombuses are there made
up of two adjacent small triangles?
Solution:
? 18
8
[TMC Regional 2008 Q3] In total how
many triangles of any size are there in the
diagram?
Solution: 35
?
Solution:
? 85
10 [JMO 1997 A8] Given a cube, each
selection of three vertices 𝐴, 𝐵, 𝐶
produces a triangle. How many of
these triangles are right-angled
triangles? (Triangles may only be on
the faces of the cube)
Solution: 24
?
Exercise 1
11 Each hour a pirate ship can either sail
1km West (W), 1km East (E), 1km
North (N) or 1km South (S). It must
always be moving. List all the ways in
which the ship can end up 2km North
after 4 hours (e.g. ‘NNWE’) and count
the number of possible journeys.
(Hint: it may help to break the
problem down possible sets of four
movements, e.g. a N, N, W and E, and
then considering the possible
orderings of each possible set).
Solution: NNWE, NNEW, NWNE,
NENW, NWEN, NEWN, WNNE, ENNW,
? EWNN,
WNEN, ENWN, WENN,
SNNN, NSNN, NNSN, NNNS (16 ways).
Exercises on provided sheet.
12
[TMC Regional 2014 Q4] Every day Keith
has a breakfast, a lunch and a dinner.
The options for each meal are:
One day Keith eats four items. In how
many different ways can he do this?
Solution: 12
?
13 [JMO 2004 A10] The Famous Five have
been given 20 sweets as a reward for
solving a tricky crime. They have agreed
that the oldest of them must receive
more than the next oldest, who must
receive more than the next oldest, and
so on. Assuming that each of the five
gets at least one sweet, in how many
different ways can they share the
sweets?
Solution:
? 7
Exercise 1
14
Exercises on provided sheet.
[Cayley 2007 Q4] How many right-angled 15 [IMC 2009 Q22] A square is divided
triangles can be made by joining three
into eight congruent triangles, as
vertices of a cube? (This time triangles
shown. Two of these triangles are
formed may also go inside the cube)
selected at random and shaded black.
Solution: 48 (In a previous question we
What is the probability that the
found 24 triangles on the faces. A further
resulting figure has at least one axis of
24 triangles can be formed, 2 for each
symmetry?
1
4
1
face diagonal where?the triangle will pass
A 4
B 7
C 2
inside the cube, e.g:).
5
D 7
E 1
Solution: D
14
There are 5 people in a room. How many
ways are there of forming two teams of
people, where each team must have at
least one person.
Solution: 15 teams. (5 teams where the
teams are split 1 and 5. 10 teams where
?
the teams are split 2 and 3)
?
PART 2 :: Counting by Multiplying
Starter: Suppose the alphabet was limited to just A, B, C, D. Fold your provided piece
of paper into 16 parts. Put ‘A’ 4 times, ‘B’ 4 times, ‘C’ 4 times and ‘D’ 4 times, before
cutting. Use these cards (or otherwise) to answer the following questions:
Note: a ‘word’ is any possible sequence of characters; it need not be in a dictionary.
Question 1
Question 2
Question 3
The number of possible
2 letter ‘words’ where
the letters are different.
The number of possible 2
letter ‘words’ where the
letters are unrestricted.
4 possibilities for first
letter, 3 for?the second.
𝟒 × 𝟑 = 𝟏𝟐
4 possibilities for first
letter, 4 for the
? second.
𝟒 × 𝟒 = 𝟏𝟔
The number of possible 3
letter ‘words’ where each
letter cannot be the same
as the previous letter, but
is otherwise unrestricted.
Question 4
Question N
Question NN
The number of 4 letters
words using each of the
letters A, B, C, D exactly
once.
The number of 5 letter
words using the letters
AABCD, each exactly once.
The number of 10 letter
words using the letters
AAAAAABBCD each once.
𝟒 × 𝟑 × 𝟐?× 𝟏 = 𝟐𝟒
𝟓×𝟒×𝟑×𝟐
? = 𝟔𝟎
𝟐
𝟒 × 𝟑 ×?𝟑 = 𝟑𝟔
𝟏𝟎!
=
? 𝟐𝟓𝟐𝟎
𝟔! 𝟐!
Multiplying and Factorial Function
If the number of choices for each ‘thing’ are independent of each other, then we can
multiply them together to get the total number of combinations.
ABCDEFGHIJKLMNOPQRSTUVWXYZ
For a full alphabet of 26 letters:
a) How many possible 3 letter ‘words’ are there?
There are 26 choices for each letter. ?
𝟐𝟔 × 𝟐𝟔 × 𝟐𝟔 𝒐𝒓 𝟐𝟔𝟑
b) How many possible 3 letter ‘words’ are there, where all letters are different?
Each time there is one less choice.
𝟐𝟔 × 𝟐𝟓 × 𝟐𝟒
?
c) How many possible ways are there of arranging the letters of the word MATHS?
There are 5 choices for the first letter, then 4 for the next, and so on.
?
𝟓×𝟒×𝟑×𝟐×𝟏
! 5! = 5 × 4 × 3 × 2 × 1 is said “5 factorial”.
𝑛! means the number of ways of arranging 𝑛 distinct
objects in a line.
Examples
Q
I throw 4 dice. How many possible outcomes are there?
𝟔 × 𝟔 × 𝟔 × 𝟔 𝒐𝒓 𝟔𝟒
?
Q
I throw 10 coins. How many possible outcomes are there?
𝟐𝟏𝟎 = 𝟏𝟎𝟐𝟒
?
Q
I have a 10 kittens. I pick up 4 kittens and put them in a
line. How many possibilities are there?
𝟏𝟎 × 𝟗 × 𝟖 × 𝟕 = 𝟓𝟎𝟒𝟎
Note that the ordering of kittens in the line matters in
?
this scenario (the number of possibilities
would be less if
we only cared about what kittens were selected)
A Few More Examples
Q
I have 6 different coloured balls in a line. How many ways of arranging them?
𝟔! = 𝟕𝟐𝟎
?
Q
3 boys and 4 girls go to a cinema. How many ways of arranging them such that
the boys all come before the girls?
𝟑! × 𝟒! = 𝟏𝟒𝟒
?
Q
How many ways of arranging the letters in the word BANANA?
If the letters were all different there would be 𝟔! = 𝟏𝟐𝟎 ways.
However, 𝑩𝑨𝑨𝑨𝑵𝟏 𝑵𝟐 for example would count as the same as 𝑩𝑨𝑨𝑨𝑵𝟐 𝑵𝟏
because the N’s aren’t distinguishable. So we need to divide by 2 to avoid
duplicates. Similarly there are 𝟑! = 𝟔 ways
? the A’s could have been arranged
which would lead to the same possibility. Thus:
𝟔!
= 𝟔𝟎
𝟐! 𝟑!
Test Your Understanding
1
A pack of cards contains 52 cards, 13 of each of four suits (Hearts, Spades, Clubs,
Diamonds). In each suit we have the cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack,
Queen, King.
Determine the number of possibilities when:
a I deal out 3 cards and put them in a line:
𝟓𝟐 × 𝟓𝟏 ?
× 𝟓𝟎
b I deal out Ace, 2, 3 and 4 of Diamonds, them arrange them in a line.
𝟒! = 𝟐𝟒 ?
c I put 6 cards in a line, the first three Jack, Queen and King of Clubs, and the
second three Jack, Queen and King of Spades.
𝟑! × 𝟑! =?𝟑𝟔
d I deal out all the spades them arrange them in a line.
𝟏𝟑! = 𝟔 𝟐𝟐𝟕
? 𝟎𝟐𝟎 𝟖𝟎𝟎
e I deal out all the cards and put them all together, but ensure the cards in
each suit are kept together.
There are 𝟏𝟑! ways or arranging the clubs, 𝟏𝟑! ways of arranging the
spades, etc. But there’s also 4! ways?of arranging the four blocks of cards.
𝟏𝟑! × 𝟏𝟑! × 𝟏𝟑! × 𝟏𝟑! × 𝟒! = 𝒂 𝒃𝒊𝒈 𝒏𝒖𝒎𝒃𝒆𝒓
2
a
b
c
How many ways of arranging the letters of:
PLUM
𝟒! = 𝟐𝟒 ?
APPLE
𝟓! ÷ 𝟐 = 𝟔𝟎
?
RASPBERRY
𝟗! ÷ 𝟑! = ?
𝟔𝟎 𝟒𝟖𝟎
Exercise 2
Exercises on provided sheet.
1 A queue in the post office consists of a cat, a dog
and a llama. How many possible orderings are
there of animals in the line?
Solution: 𝟑! = 𝟔
?
2 A bag contains 8 different coloured balls. I pick 4 of
them and put them in a line. How many ways are
there of doing this?
Solution: 𝟖 × 𝟕 × 𝟔 × 𝟓 = 𝟏𝟔𝟖𝟎
?
4 How many ways are there of arranging the
letters in the words:
a. SMURF
𝟓! = 𝟏𝟐𝟎
b. BOTTLE
𝟔! ÷ 𝟐! = 𝟑𝟔𝟎
c. CABBAGE 𝟕! ÷ 𝟐! 𝟐! = 𝟏𝟐𝟔𝟎
d. TITILLATION 𝟏𝟏! ÷ 𝟑! 𝟐! 𝟐! = 𝟓𝟓𝟒𝟒𝟎𝟎
?
?
?
?
5 [TMC Regional 2009 Q1] A tile is fixed to a wall
and then painted with four different colours,
one for each quarter. One way of doing this is
3 I have cards with the number 1, 2, 3, 4 on them. I
have a plentiful supply of each. How many ways
are there of:
In how many other ways may the tile be
a. Putting four numbers in a line where all
painted?
𝟒! − 𝟏 = 𝟐𝟑
numbers must be different.
𝟒! = 𝟐𝟒
b. Putting four numbers in a line where we are
[JMO 1999 A5] UKMT, TMUK and KTUM are all
unrestricted in choices of numbers. 𝟒𝟒 = 𝟐𝟓𝟔 6 different arrangements of the letters U, K, M
c. Putting four numbers in a line where each
and T. If the number of all the different
number must not be the same as the one
arrangements of these four letters is 𝑝 and the
immediately before.
𝟒 × 𝟑𝟑 = 𝟏𝟎𝟖
number of all the different arrangements of the
d. Forming a four digit number starting with 3.
letters 𝑈, 𝐾, 𝐽, 𝑀 and 𝑂 is 𝑞, what is the value of
𝟑
𝟒 = 𝟔𝟒
𝑞
𝟓!
𝟓×𝟒!
?
Solution: =
=𝟓
?
?
?
?
?
𝑝
?
𝟒!
𝟒!
Exercise 2
7
Exercises on provided sheet.
[JMC 2013 Q25] For Beatrix’s latest art installation,
she has fixed a 2 × 2 square sheet of steel to a wall.
She has two 1 × 2 magnetic tiles, both which she
attaches to the steel sheet, in any orientation, so
that none of the sheet is visible and the line
separating the two tiles cannot be seen. As shown
alongside, one tile has one black cell and one grey
cell; the other tile has one black cell and one spotted
cell.
How many different looking 2 × 2 installations can
Beatrix obtain?
A 4
B 8
C 12
D 14
E 24
Solution: C
The dotted tile can go in 4 positions. This leaves 3
positions for the grey tile. The rest are filled with
black (with no choice involved). 𝟒 × 𝟑 = 𝟏𝟐
?
8
[TMC Regional 2013 Q8] Claire,
David, Jean and Richard are queuing
for the bus. In how many different
ways can they line up in single file,
one behind the other, without Jean
being last?
?
Solution: Starting from the back
𝟑 × 𝟑 × 𝟐 × 𝟏 = 𝟏𝟖
9 [Kangaroo Pink 2015 Q18] Petra has
three different dictionaries and two
different novels on a shelf. How
many ways are there to arrange the
books if she wants to keep the
dictionaries together and the novels
together?
A 12
B 24
C 30
D 60
E 120
Solution: B (Can either have booksnovels or novels-book, giving 2 lots
of 𝟑! × 𝟐!)
?
Exercise 2
10 [IMC 2015 Q20] A voucher code is made
up of four characters. The first is a letter:
V, X or P. The second and third are
different digits. The fourth is the units
digit of the sum of the second and third
digits. How many different voucher codes
like this are there?
A 180 B 243 C 270
D 300 E 2700
Solution: C. 3 choices for first character.
10 choices for second character and 9 for
third (since a different digit). There is no
element of choice for the fourth
?
character as it is determined by the
second and third.
𝟑 × 𝟏𝟎 × 𝟗 = 𝟐𝟕𝟎
Exercises on provided sheet.
11
[IMC 2008 Q23] Beatrix has a 24-hour
digital clock on a glass table-top next
to her desk. When she looked at the
clock at 13:08, she noticed that the
reflected display also read 13:08, as
shown. How many times in a 24-hour
period do the display and its reflection
give the same time?
A 12
B 36
C 48
D 72
E 96
Solution: E. First note that 0, 1, 3 and
8 reflect.
?
𝟐 × 𝟒 × 𝟑 × 𝟒 = 𝟗𝟔
Exercise 2
12
Exercises on provided sheet.
[Kangaroo Grey 2006 Q16/Pink 2006 Q19] A train consists of five
carriages: I, II, III, IV and V. In how many ways can the carriages be
arranged so that carriage I is nearer to the locomotive than carriage II is?
A 120 B 60
C 48
D 30
E 10
Solution: B
Long way:
I is in 1st position: 𝟒! = 𝟐𝟒 ways of arranging remaining carriages.
I is in 2nd position: 𝟑 × 𝟑! = 𝟏𝟖 ways
?
I is in 3rd position: 𝟑 × 𝟐 × 𝟐 = 𝟏𝟐 ways
I is in 4th position: 𝟑! = 𝟔 ways
𝟐𝟒 + 𝟏𝟖 + 𝟏𝟐 + 𝟔 = 𝟔𝟎 ways
Really Short way: In half of the 𝟓! = 𝟏𝟐𝟎 ways of arranging the five
carriages, I will appear before II. 𝟏𝟐𝟎 ÷ 𝟐 = 𝟔𝟎
The ‘Choose’ Function
𝐵1
𝐵2
𝐵3
𝐵4
𝐵5
How many ways are there of selecting 2 balls from the 5?
(such that the order of the balls in my selection does not matter)
There are 5 balls for our first choice.
There are then 4 balls to choose from.
This initially gives 20 possibilities.
?
However, 𝐵1 𝐵4 for example would represent the same choice as 𝐵4 𝐵1 .
We therefore have to divide by 2 to avoid duplicates.
There are 10 possible selections.
𝑛
! 𝑛𝐶𝑟 (often written
), said “𝑛 choose 𝑟”, is the number of ways of
𝑟
‘choosing’ 𝑟 items from 𝑛 without duplicates, such that order does not matter.
𝑛!
𝑛
=
𝑟
𝑟! 𝑛 − 𝑟 !
Practise using the formula (and your calculator)
9!
9
?= 84
=
3
3! 6!
10!
10
?= 10
=
1
1! 9!
Broculator Tip: Use the
𝑛𝐶𝑟 button on your
calculator (SHIFT ÷) to
calculate directly.
4!
4
?= 1
=
0
0! 1!
What would we expect this to
be? (The number of ways of
choosing 0 items from 4)
What would we expect this to
be? (The number of ways of
choosing 1 item from 10)
Note that ‘no selection’
is itself a choice!
Bro Side Note: The
choose formula only
works if 0! = 1
Examples
Q In the UK National Lottery, you pick 6 numbers out of 59.
How many possible lottery tickets are there?
𝟓𝟗
= 𝟒𝟓 ?
𝟎𝟓𝟕 𝟒𝟕𝟒
𝟔
Q Bob the Teacher needs to form a maths team for a competition.
He needs 3 boys and 3 girls for the team. He has 10 boys and 10
girls to choose from. How many possible teams are there?
𝟏𝟎
𝟏𝟎
×
= 𝟏𝟒 𝟒𝟎𝟎
𝟑
𝟑?
Test Your Understanding
A Bob the Football Manager needs to select 11 players from his
squad of 15 players. How many possible selections are there?
𝟏𝟓
= 𝟏𝟑𝟔𝟓
?
𝟏𝟏
B I have a bowl of 5 pieces of fruit, containing an orange, apple, banana, grapefruit and
kiwi. I also have in my cupboard 4 chocolate bars: a Kit Kat, Snickers, Mars Bar and Twix.
If I want two pieces of fruit and two chocolate bars for my lunch, how many possible
selections are there?
𝟒
𝟓
×
= 𝟔𝟎
𝟐?
𝟐
N In this grid of 16 squares, I choose 4 to colour red followed by 4 to
colour yellow. Determine the number of possible colourings.
𝟏𝟔
𝟏𝟐
×
?= 𝟗𝟎𝟎𝟗𝟎𝟎
𝟒
𝟒
Exercise 2
1 I am building a car park and need to choose 2
of the 20 spaces to be disabled parking spaces.
How many ways are there of doing this?
Solution: 190
?
2
In a lottery I can choose 4 distinct numbers
between 1 and 20. How many possible lottery
tickets are there? Solution:?4845
3 Of 9 balls in a line, 5 need to be painted red
(and those not chosen will be painted blue).
How many ways are there of doing this?
Solution: 126
?
4 Of 9 balls in a line, 4 need to be painted blue
(and those not chosen will be painted red).
a. How many ways are there of doing
this? Solution:
? 126
b. Hence, what other ‘choose’ will have
50
𝟓𝟎
the same value as
? Sln: ?
20
𝟑𝟎
Exercises on provided sheet.
5 Alice is asked to think of 3 different
numbers between 1 and 8, and Bob
is asked to do the same. How many
total possible combinations are
there for Alice and Bob’s selection
combined?
Solution: 𝟓𝟔 × 𝟓𝟔 = 𝟑𝟏𝟑𝟔
?
6 As part of winning a car, I have to
choose 4 numbers between 1 and
15 and throw three dice. How many
total possible outcomes are there?
Solution: 𝟏𝟑𝟔𝟓 ?
× 𝟔𝟑 = 𝟐𝟗𝟒𝟖𝟒𝟎
7
I choose 4 distinct numbers
between 1 and 8. How many
possibilities are there where the
numbers in my selection add to give
an even number?
Solution: It will be
? half of them. 35
Exercise 2
8
0
on a
0
row in your book. On the next row
1
1
write
and
. On the next row
0
1
2
2
2
write
,
and
. Continue
0
1
2
this pattern for a few more rows,
where the top number increases by 1
for each row, and the bottom
number varies between 0 and the
top number.
a. What do you notice about
the numbers you get in your
structure?
You get Pascal’s
? Triangle!
b. Add up the numbers in each
row (your first two totals
should be 1 and 2). What do
you notice?
You get powers
? of 2!
Use your calculator to write
Exercises on provided sheet.
9
A beaver travels from the bottom-left hand
corner to the top-right corner of a 4 × 4
grid, each time only making up or right
movements. One possible paths is as
pictured.
a. How many possible paths are
there?
Solution: Of the 8 moves 4 need to
𝟖
be right moves.
= 𝟕𝟎
𝟒
?
(Hint: write a few possible paths as a
sequence of arrows, e.g. →↑→→↑↑→↑.
What’s always the same about such valid
sequences of moves?)
b. If the beaver is travelling across a
𝑛 × 𝑛 grid, write an expression for
the total number of paths in terms
𝟐𝒏
of 𝑛. Solution:
𝒏
Finish
?
Start
Exercise 2
N
Exercises on provided sheet.
I choose 5 numbers between 1 and 20. How many choices are
there such that the range of the numbers in my selection is 6?
Solution: Suppose the smallest number was 1 and the largest 7.
Then the remaining three numbers can be chosen between 2
𝟓
and 6, giving
= 𝟏𝟎 possibilities.
𝟑
?
We’ll have the same number of possibilities where the smallest
and largest numbers are 2 and 8, and so on, up to 14 and 20.
That’s 𝟏𝟒 × 𝟏𝟎 = 𝟏𝟒𝟎 possibilities.
Levelled Activity
Level 3
Level 2
Level 1
(Teacher Note:
See printout)
Instructions: You may work in pairs. Start on the Level 1 problems
provided, recording your answer on this sheet. Answers to Level 1
questions are at the front of the class. Once you’ve checked your
answers, go to your teacher and ask for Level 2.
You will need to ask the teacher to check your answers to Level 2 and
beyond.
Solutions are on the next slides.
Level 1 Solutions
1 How many possible 3 letters ‘words’ are there
using letters from the English alphabet.
𝟑 = 𝟏𝟕𝟓𝟕𝟔
Solution: 𝟐𝟔?
2 I take 4 items from 7 and put them in a line.
How many possibilities are there?
Solution: 𝟕 × 𝟔?× 𝟓 × 𝟒 = 𝟖𝟒𝟎
3 From 9 people I choose 4 people to form a
team. How many possible teams are there?
Solution:
?𝟗𝟒
= 𝟏𝟐𝟔
4 How many squares of any size are in this
diagram?
Solution: 𝟏𝟔 + 𝟗 + 𝟒 + 𝟏
?
= 𝟑𝟎
5 How many ways can I arrange 6
books on a shelf?
Solution: 𝟔!?= 𝟕𝟐𝟎
6 How many ways can you arrange the
letters of the word CATTLE?
𝟔!
Solution: 𝟐!?= 𝟑𝟔𝟎
7 [JMC 2000 Q11] The digits of this
year, 2000 A.D., add up to 2. In how
many other years since 1 A.D. has
this happened? Solution:
? 9
8 [JMC 2007 Q7] The equilateral
triangle 𝑋𝑌𝑍 is fixed in position. Two
of the four small triangles are to be
painted black and the other two are
to be painted white. In how many
different ways can this be done?
𝟒
Solution:?
=𝟔
𝟐
Level 2 Solutions
1 How many possible combinations of
outcomes are there from the throw of
three dice and five coins?
Solution: 𝟔𝟑 ×?𝟐𝟓 = 𝟔𝟗𝟏𝟐
2 [JMO 2006 A4] An equilateral triangle is
drawn on a sheet of white card and
divided into three identical regions as
shown. Then each region is painted red
or yellow or blue. More than one region
may be painted in the same colour. How
many different painted triangles can be
made in this way? (Rotating a triangle
does not make it different.)
Solution: 11
?
3 [JMO 2008 A4] How many three-digit
numbers have the product of their digits
equal to 6?
Solution: 9?
4 [TMC Regional 2012 Q9] In how
many ways can you split your team
of 4 people into two separate
teams, so that there is at least one
person in each team?
Solution: 7?ways
5 [Kangaroo Grey 2009 Q16] How
many ten-digit numbers are there
which contain only the digits 1, 2 or
3, and in which any pair of adjacent
digits differs by 1?
A 16
B 32
C 64
D 80
E 100
Solution: C.
If the number starts with 2, the next is 1
or 3, the next is 2, and so on. When we
don’t have a 2 we have two choices, so
𝟐𝟓 = 𝟑𝟐.
But we could had 2 second, giving
another 32 possibilities.
?
Level 3 Solutions
1
[Cayley 2008 Q1] How many four-digit multiples
of 9 consist of four different odd digits?
Solution: 24. Digits have to add up to multiple of
9: only possibility for digits is 1359. There’s 4!
ways of arranging these.
?
2
[IMC 2010 Q24] A new taxi firm needs a
memorable phone number. They want a number
which has a maximum of two different digits.
Their phone number must start with the digit 3
and be six digits long. How many such numbers
are possible?
A 288
B 280
C 279
D 226
E 225
If other digit is different 0, then there are 2
choices for each of the 5 other digits, giving 𝟐𝟓 =
𝟑𝟐 possibilities. However, we want to exclude
the case where they’re all 3s (for the moment),
giving 31 possibilities There’s 9 numbers this
other digit could be giving 𝟗 × 𝟑𝟏 = 𝟐𝟕𝟗. Then
we count the one possibility where all the digits
are 3, giving 280.
?
3 [SMC 2005 Q16] A hockey team
consists of 1 goalkeeper, 4
defenders, 4 midfielders and 2
forwards. There are 4 substitutes: 1
goalkeeper, 1 defender, 1
midfielders and 1 forward. A
substitute may only replace a
player of the same category, e.g.
midfielder for midfielders. Given
that a maximum of 3 substitutes
may be used and that there are still
11 players on the pitch at the end,
how many different teams could
finish the game?
A 110 B 118 C 121
D 125 E 132
Solution:?
B
Level 3 Solutions
6 [Kangaroo Pink 2007 Q24] At a party, five
4 [JMO 2004 A10] The Famous Five have been
girls give each other gifts in such a way
given 20 sweets as a reward for solving a tricky
that everybody gives one gift and
crime. They have agreed that the oldest of
everybody receives one (though of
them must receive more than the next oldest,
course nobody receives their own gift).
who must receive more than the next oldest,
How many possible ways are there for
and so on. Assuming that each of the five gets
this to happen?
at least one sweet, in how many different ways
A 5
B 10
C 44
can they share the sweets?
D 50
E 120
Solution: 7
Solution: C. Suppose the girls are ABCDE.
?
5
[JMO 2002 B1] A number like 4679 is called an
ascending number because each digit in the
number is larger than the preceding one.
(i) How many ascending numbers are there
between 1000 and 2000?
(ii) How many ascending numbers are there
between 1000 and 10000?
Solution: 56?and 126
Then to avoid giving themselves a
present, we either have to have a ‘cycle’
of 3 and 2 (e.g. ABC give each other
presents, and DE give each other one) or
𝟓
a ‘cycle’ of 5. There are
= 𝟏𝟎
𝟐
possible groupings for 2-3 cycles, where
for each there’s only 2 possible giving of
presents, giving 20 possibilities. If the
cycle is 5, then there’s 𝟒! = 𝟐𝟒
possibilities (as A had 4 people he can
give his present to, then B has 3, etc.)
?
Level 3 Solutions
7
[SMC 2001 Q21] A postman’s sack contains five letters, one each for the five houses
in Cayley Close. Mischievously, he posts one letter through each door without looking
to see if it is the correct address. In how many different ways could he do this so that
exactly two of the five houses receive the correct letters?
A 5
B 10
C 20
D 30
E 60
𝟓
= 𝟏𝟎 ways in which we could choose 2 houses to receive the correct
𝟐
letter. Then of the remaining 3 houses, there are only 2 ways in which they could
receive the wrong letter (if the houses are?ABC, then the letters could be BCA or
CAB).
There are
That gives 𝟏𝟎 × 𝟐 = 𝟐𝟎 ways.