Combinatorics Slides

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Topic 4: Combinatorics
Dr J Frost ([email protected])
Last modified: 4th August 2013
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The level, 1 being the easiest, 5
the hardest, will be indicated.
BMO
MAT
British Maths Olympiad
Frost
A Frosty Special
Questions from the deep dark
recesses of my head.
Those with high scores in the SMC
qualify for the BMO Round 1. The
top hundred students from this go
through to BMO Round 2.
Questions in these slides will have
their round indicated.
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Computer Science at Oxford
University.
University Interview
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Well known problems in maths.
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
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Question: The capital of Spain is:
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
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
C: Madrid

Topic 4: Combinatorics
Unit 1 - Fundamentals
a. Slot Filling
b. The Factorial Function – n!
c. The Permutation Function – nPk
d. The Choose Function – nCk
e. Distinguishable vs Indistinguishable Objects
f. Ordered vs Unordered Objects
g. Probabilistic vs Combinatoric Approaches
Topic 4: Combinatorics
Unit 2 – More on Counting Problems
a. Converting Problems into Slot-Filling Ones
b. Purposely Overcounting
c. Typed Problems
d. Objects in a Ring
e. Currency Problems
f. Multi-Step problems
Topic 4: Combinatorics
Unit 3 – Recurrence Relations
a. Identifying the State and Actions
b. Forming the Recurrence
c. Dynamic Programming
Unit 4 – Compositions and Partitions
a. Compositions
b. Partitions
Epilogue – Geometric Arrangements
Topic 4 – Combinatorics
Part 1: Fundamentals
Here, we look at the some of the basics of combinatorics, and
fundamental combinatoric operators that are commonly used.
What is combinatorics?
Broadly, the ‘number of ways’ of arranging
things according to some structure and
constraints.
“How many ways are there of matching 3
numbers on the lottery?”
“How many ways of making a selection of
fruit from a bowl of 10 items?”
“How many ways can 10 people in a circle
shake hands simultaneously such that no
handshake crosses?”
This is closely related to probability, since we often have to consider combinations and
permutations of things to determine the exact probability. But combinatorics arises in many
mathematical fields – you may have encountered it in Algebra for Binomial Expansion.
Fundamentals #1: Slot Filling
When considering the number of possible combinations, it’s
often helpful to think of ordered ‘slots’ in a line that we can fill.
Example: How many outcomes are there when we simultaneously roll 5 dice?
6
x 6
x
6
x 6
x
6
There are 5 slots for the 5 dice. In each slot,
there are 6 outcomes.
We just multiply the possible outcomes in
each slot to get the total number of
combinations.
= 65
Fundamentals #1: Slot Filling
When considering the number of possible combinations, it’s
often helpful to think of ordered ‘slots’ in a line that we can fill.
a) The number of ways of picking 1 card from each
of 20 different packs (of standard playing cards)?
52?20
(Click question marks
to reveal answer)
b) The number of outcomes from the throw of a
die and the toss of a coin?
6 x 2?= 12
c) The number of outcomes from 5 coins and 5
dice flipped/thrown simultaneously?
25 x 65? = 125
Fundamentals #2: The Factorial Function
Perhaps the most key operator in combinatorics is the factorial
function, where n! means “n factorial”. It tells us the number
of ways of arranging distinct objects in a line.
Q: How many ways of ordering these pieces of fruit?
We use a slot filling approach:
4
We could pick any of the 4 pieces
of fruit for the first slot.
x 3
x
2
x
We have 3 remaining fruits to
choose from for the second slot.
= 4!
1
And so on
Fundamentals #2: The Factorial Function
Puzzle: If I throw six dice simultaneously, what’s the probability that I
get a full run of 1, 2, 3, 4, 5, 6 (where the ordering of the dice does
not matter).
p(full run) = 6!?6
6
Hint: Recall from basic probability that:
p(event) = outcomes in which this event occurs
total outcomes
The particular point of interest here is that we’ve turned an unordered problem (i.e. the ordering
of the dice doesn’t matter) into an ordered one (i.e. we considered ‘ordered slots’). We’ll consider
unordered vs ordered problems later.
Question: Simplify  − 1 !
!
−1 !
−1 × −2 ×⋯

=
=
!
 ×  − 1?×  − 2 × ⋯ 
Fundamentals #2: The Factorial Function
Formal definition of factorial function:
0! = 1
(n+1)! = (n+1)n!
Base case
Recursive case
(For example
4! = 4 x 3! )
(Note that by combining the above, 1! = 1 x 0! = 1 x 1 = 1)
Just for your interest...
The factorial function works only on non-negative integers. There’s a function called the gamma function
that generalises factorial to all numbers (including decimal, negative and complex numbers!). The gamma
function is offset from the factorial function by 1. It is used for example in Number Theory and Statistics.
Г(6) = 5!
Г(22) = 21!
Г(i) = -0.1549
Г(3.5) = 3.3234 (which is between 2! and 3!)
Fundamentals #3: The Permutation Function nPk
Before, we were interested in how to order n things in a line.
But suppose now that we’re interested in how to order k things in a
line, when we can choose from n objects.
Example: How many ways to pick 3 items of fruit from amongst 5,
and put them on a shelf?
5
We could pick any of the 5 pieces
of fruit for the first slot.
x 4
x
3
= 5!
2!
= 5P3
The difference from earlier is that we only have
3 slots, and thus don’t use all the fruit.
Fundamentals #3: The Permutation Function nPk
The cards in a set of 36 are numbered 1 to 36. The cards
are shuffled and four cards are dealt. What are the
chances of them being dealt in descending order?
(Click your choice of answer)
A: 1 in
2
B: 1 in
8
D: 1 
in 24
E: 1 in
 36
C: 1 in
 16
SMC
Level 5
Level 4
Notes: If 4 (distinguishable) objects are chosen, there
are 4! = 24 ways to arrange them. Only one of these
possibilities has them in descending order.
Level 3
Level 2
Level 1
Fundamentals #3: The Permutation Function nPk


!
=
− !
It gives us the number of ways
of putting n items into k slots.
Like the factorial function, you can find this on your calculator (usually
you need to press the ‘2nd Function’ button first).
This combinatoric function tends to be seen less frequently. But we can
modify it to give us something much much more common...
Fundamentals #4: The ‘Choose’ Function nCk
Suppose again we pick k items from n, but now, the order
in which we pick them doesn’t matter.
With our fruit example, we might consider how many ways we can make
a selection of 3 pieces of fruit from 5.
We could use 5P3 to give the number of ways of picking 3 pieces of fruit
from 5, where the order in which we picked them mattered.
But we wish to disregard the order: for example, Orange-Lemon-Lime is
equivalent to a selection of Lemon-Lime-Orange. Thus each 3!
possibilities only represents one unordered selection. Thus:

Ways of making a selection of 3 fruits
=

!
=
!
!!
Fundamentals #4: The ‘Choose’ Function nCk
This is known as the ‘choose’ function, because it gives us the number of
ways of ‘choosing’ k items from n.
(You can again find it on a standard scientific calculator)
nC
We rarely write
nC
k
=
__n!__
(n-k)!k!
n
k in practice – instead we’d write ( k
)
Question: How many possible lottery tickets are there, if there’s 49
possible numbers a ticket consists of 6 distinct numbers?
49
( 6 ) ≈ 14 million. Of the 49 numbers, 6 can be chosen for a ticket,
? matter.
where the ordering of the numbers doesn’t
Fundamentals #4: The ‘Choose’ Function nCk
You ought to remember these special cases:
(they can all be worked out using the definition of nCk)
n
?
=
1
0
( )
n
2
( )
n(n-1)
=
?
2
n
?
=
n
1
( )
n
?
=
1
n
( )
Fundamentals #4: The ‘Choose’ Function nCk
Its use in Binomial Expansion
You may have encountered the use of the ‘choose’ function as a Binomial
Coefficient – the coefficients of each term in a polynomial resulting from the
expansion of a bracket with two items in it and to some power.
(1+x)4 =
=
4 4
( 0 )x +
x4 + 4x3
4 3
4 2
( 1 )x + ( 2 )x
+ 6x2 + 4x + 1
+
4
( 3 )x
+
4
( 4 )1
If we consider all the brackets and what happens when we expand, how many x2
terms will we see? Recall that to expand brackets, we consider all possible choices
when we pick 1 item from each bracket.
These four items multiplied
together will give an x2 term.
(1+x)4 = (1+x)(1+x)(1+x)(1+x)
Similarly, these will give an x2
term.
Fundamentals #4: The ‘Choose’ Function nCk
Its use in Binomial Expansion
(1+x)4 = (1+x)(1+x)(1+x)(1+x)
There’s 4 brackets from which we could choose the two x terms needed to make x2
(and from the brackets we don’t choose from, we use a 1), so we’ll end up with ( 42 )
x2 terms in the expansion!
#5: Distinguishable vs Undistinguishable
Consider 3 blue balls and 4 orange. How many ways are there of arranging
them in a line?
If the balls are all
distinguishable...
Then we can tell them all apart. And thus, there’s
simply 7! ways of arranging them.
If the balls are
indistinguishable...
Then we can’t tell balls of the same colour apart.
Then the problem is slightly more complicated...
Sometimes, a question will explicitly state whether objects are distinguishable or not.
If not, then use your common sense. For this particular example, the fact it’s
mentioned groups of balls that share a property (i.e. colour) suggests you can’t tell
them apart within that group.
#6: Ordered vs Unordered
Recall that ‘ordering’ relates to whether the order in which the items are
selected matter. The ‘choose’ function was the unordered cousin of the
‘permutation’ function for example.
Sometimes problems have multiple approaches:
Question: Calculate the probability of winning the UK lottery, where you
choose 6 distinct numbers from 1 to 49 for your ticket, and 6 numbers are
drawn from the machine (i.e. Ignoring the ‘bonus ball’).
a) The unordered approach
If at no point do we consider the balls ordered, then:
Total possible tickets
Winning outcomes
49C
6
1 (because only one unordered choice of numbers can win)
p(win) =
1_
49C
6
_
#6: Ordered vs Unordered
Question: Calculate the probability of winning the UK lottery, where you choose 6 distinct
numbers from 1 to 49 for your ticket.
b) The ordered approach
Consider the numbers on a lottery ticket as ordered (since the balls from the machine
are drawn in a particular order!). Then:
Total possible tickets
Winning outcomes
49P
6
(because we’re considering ordering)
6! (because if the winning numbers were 13-3-9-40-24-1 in the
order that they were drawn, then any of the 6! arrangements of
these would win)
p(win) =
_6!_
49P =
6
_6!_
49!/43!
=
_43!6!_
49
=
_1_
49C
6
These approaches are subtly different, even if they end up with the same probability.
While in this particular case, approach (a) seems simpler, in combinatoric problems where
we have a mixture of ordered and unordered items, or distinguishable and
indistinguishable, then the latter ‘ordered’ approach might be required.
#7: Probabilistic vs Combinatoric Approaches
Question: 3 men each independently pick a number between 1 and
10. What’s the probability that their numbers are different?
4
6
_18_
p(different numbers) = ?
25
1
#7: Probabilistic vs Combinatoric Approaches
The probabilistic approach
The combinatoric approach
Consider the men one at a time,
and consider the probability that
their number doesn’t clash with the
previous men’s choice.
Total outcomes:
103
Note that this is the total number of
ordered choices.
Because there’s no
previous men!
Hammond goes first:
p(number doesn’t clash) = 1
9 of the 10 numbers
won’t be the same
as Hammond’s
May goes second:
p(number doesn’t clash) = 9
10
Outcomes in which numbers are
different:
The number of ways of choosing 3
numbers from 10 (but to be
consistent, we have to consider their
order), i.e. 10P3
Clarkson goes last:
p(number doesn’t clash) = 8
10
p(all numbers different) =
p(all numbers different) = 1 x 9 x 8 = 18
10
10
25
(We’re of course making an independence assumption here: i.e. the men’s
choices of numbers didn’t influence the others.)
10
18
=
103 25
3
Note: Whenever we want to consider distinct
choices of items, you should always immediately
think to use the ‘choose’ function.
Topic 4 – Combinatorics
Part 2: More on Counting Problems
Here, we drill in to the types of counting problems that are
frequently seen, and strategies to deal with each one.
Tip #1: Converting problems into slot filling ones
You have a bowl of 3 pieces of fruit. How many ways are there of
making a selection of fruits from the bowl (you can include the
possibility that you pick no fruit).
Answer = 8?
Tip #1: Converting problems into slot filling ones
The ‘choose’ approach
The slot filling approach
In our selection, we can just choose 1
piece of fruit:
( 31 ) = 3
Or we can choose 2 pieces of fruit:
( 32 ) = 3
Or we can choose all 3 pieces of fruit:
( 33 ) = 1
Give each piece of fruit a ‘slot’.
Each fruit can either:
a) Be chosen
b) Not be chosen
Or choose none of them:
Thus each slot has 2 possible states.
( 30 ) = 1
This gives 23 = 8 possibilities.
This approach is clearly much
simpler!
Tip #1: Converting problems into slot filling ones
The fact either approach to this fruit
problem gives the same answer yields a
nice identity!
Using ‘choose’
approach for n fruits.
Using ‘slot based’
approach for n fruits.
This identity can also be seen by looking at the sum of each row in Pascal’s triangle:
2
0
2
1
2
2
( ) ( ) ( )
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Sum = 1 = 20
Sum = 2 = 21
Sum = 4 = 22
Sum = 8 = 23
Tip #1: Converting problems into slot filling ones
How many ways of making a selection of fruits from 5 apples and 3
oranges?
Answer = 24
This time, anything other than a ‘slot based’
approach will cause a massive amount of
headache.
Have a slot for the number of apples, and a slot
for the number of oranges.
We can have 0-5 apples (6 possibilities) and 0-3
oranges (4 possibilities).
?
So 6 x 4 = 24 possibilities (or 23 if we exclude
the case where we pick no fruit)
Tip #2: Purposely overcounting
Sometimes it’s easier to consider more possibilities than we initially need
(because the result is easy to calculate) before eliminating possibilities we
don’t want.
Question: How many ways can we pick at least 2 different kinds of fruit from a bowl,
given there are 3 apples, 2 oranges, 1 plum and 9 limes?
x3
x2
?
Answer = 224
x1
x9
Tip #2: Purposely overcounting
Question: How many ways can we pick at least 2 different kinds of fruit from a bowl?
x3
Horrendously cumbersome approach
Work out number of ways of picking:
a) 2 types of fruit
b) 3 types of fruit
c) 4 types of fruit
And add together.
e.g. Ways of picking 2 types:
• Only apples & oranges: 3 x 2 = 6
• Only apples & plums: 3 x 1 = 3
• ...
Which gives 65 ways.
For 3 types of fruit, it’s 105 ways.
For 4 types of fruit, it’s 54 ways.
Total = 65 + 105 + 54 = 224.
x2
x1
x9
‘Overcounting’ approach
Consider ALL possible selections
first:
4 x 3 x 2 x 10 = 240
Exclude selections involving 1 type
of fruit:
3 + 2 + 1 + 9 = 15
Exclude selections involve no fruit:
1
Total = 240 – 15 – 1 = 224
Tip #3: Typed objects
Frequently, objects have certain types, e.g:
Red and blue, boys and girls, sisters and non-sisters, etc.
Question: How many ways of arranging 8 books, if you must
keep the two red ones next to each other?
Answer: 10,080
?
Explanation:
There’s 7 ways in which the two red books can appear together. Then, we can arrange the 2 red books in 2!
ways, and the 6 non-red books in 6! ways. That gives 7 x 2! x 6! = 10,080 ways.
?
Tip #3: Typed objects
The strategy for these kinds of problems is therefore:
1. Consider the possible ways in which the slots can be
assigned a particular ‘type’ (e.g. where the red books
can occur according to some constraint).
2. Multiply this by the number of ways in which the
objects can be arranged within these slots.
Tip #3: Typed objects
Here’s a harder one:
Question: 8 people go to a cinema and sit in line. 3 of them are sisters who
tend to chat if put together. How many ways are there of seating the 8 people
if the sisters aren’t allowed to sit together?
Answer: 14,400
?
Explanation:
Put the non-sisters in a line first. There’s 6 possible positions the 3 sisters can slot themselves into to avoid sitting next
to each other. That’s 6C3 = 20 possible ways we can allocate the sister seats. Then there’s 3! arrangements of sisters in
these seats, and 5! arrangements of the non-sisters in their non-sister seats. 20 x 3! x 5! = 14,400
Tip #3: Typed objects
Here’s a harder one:
Question: 8 people go to a cinema and sit in line. 3 of them are sisters who
tend to chat if put together. How many ways are there of seating the 8 people
if the sisters aren’t allowed to sit together?
BEWARE!
If we worked out the number of ways in which the sisters ARE allowed to
all sit together (i.e. 6 x 3! x 5!), you might think we can subtract this from
all possible arrangements (8!) to get the arrangements in which they
don’t sit together. But that would only give us the arrangement in which
they don’t ALL sit together, not the arrangements in which NONE of them
sit together.
Tip #3: Typed objects
Question: How many ways if arranging 3 (indistinguishable)
red balls and 5 (indistinguishable) yellow balls?
Answer: 8C3?= 56
Explanation: Observe that of the 8 slots to put the balls into, we choose 3 of them to
be red slots. Or equivalently, there’s 8! ways to arrange the 8 balls, but the red are
indistinguishable so we divide by 3!, and the yellows are indistinguishable, so we
divide by 5!
Unlike the previous problem where the sisters/red books were distinguishable, here
they are not.
Tip #4: Objects in a ring
When objects are put in the circle, this affects problems involving objects
being adjacent/not adjacent, since the line no longer has ends.
Question: There are 8 seats in a circle,
with 3 sisters. Find the number of
arrangements when the sisters:
(a) want to sit next to each
(b) don’t want to sit next to each other.
Answer (a): 5760
?
This is exactly the same as the sisters in a
line problem, except the block of 3 girls
can now be in 8 different positions, giving
8 x 3! x 5!
?
Answer (b): 7200
As before, we allocate non-sister seats
first, but now there’s 5C3 ways to allocate
the sister seats.
Tip #5: Currency Problems
A particular favourite in IMC/SMC papers is determining how many ways an
amount can be made up using certain coinage.
Question: Given an unlimited supply of 50p, £1 and £2 coins, in
how many different ways is it possible to make a sum of £100?
A: 1326

B: 2500

D: 5050

E: 10000

C: 2601

Strategy: Fix the number of say £2
coins, and then consider how many
ways there are of making up the
remaining amount using £1 and 50p
coins using the same method. Try to
spot the pattern as we increase the
number of £2 coins.
SMC
Level 5
Level 4
Level 3
Level 2
Level 1
Tip #5: Currency Problems
A particular favourite in IMC/SMC papers is determining how many ways an
amount can be made up using certain coinage.
Question: Given an unlimited supply of 50p, £1 and £2 coins, in
how many different ways is it possible to make a sum of £100?
X 20
£0 left to make
up with £1 and
50p coins.
X 19
£2 left to make up
Can have between 0
and 2 £1 coins.
3 ways.
X 18
£4 left to make up
Can have between 0
and 4 £1 coins.
5 ways.
£100 left to make up
Can have between 0
and 100 £1 coins.
101 ways.
...
X0
1 way.
Tip #5: Currency Problems
1 + 3 + 5 + ... + 99 + 101
This is an arithmetic series with a = 1, d = 2, n = 51, so Sn = ½ x 51 x (2 + 50 x 2) = 2601
Be careful about getting n right: notice that if we added 1 to all the terms and divided
by 2, we’d have the numbers 1, 2, 3, ..., 50, 51, so n = 51.
Important note: Notice that we never even had to consider the 50p coins, since by
fixing the number of £2 and £1 coins, the number of 50p coins is therefore
determined from the remaining amount. We therefore just concentrate on making
up amounts LESS OR EQUAL TO £100 with just £2 and £1 coins*, making the problem
less daunting.
* This would only not work if the remaining amount to make up with the smallest value coin is not a
multiple of the coin’s value. e.g. If the coins were just 3p and 2p and the total to make up was 15p, then
it’s not enough to just consider fixing the number of 3p coins as two and making up the rest with 2p
coins, because it can’t be done!
Tip #6: Multi-step combinatoric problems
For arrangement problems involving multiple steps:
(a) Have a clear starting point. (b) Be careful not to overcount/undercount.
Adrian teaches a class of six pairs of twins. He wishes to set up
teams for a quiz, but wants to avoid putting any pair of twins
into the same team. Subject to this condition:
i) In how many ways can he split them into two teams of six?
ii) In how many ways can he split them into three teams of
four?
Answer (i):
32 ?
Answer (ii):
960?
BMO
Round 2
Round 1
Tip #6: Multi-step combinatoric problems
For arrangement problems involving multiple steps:
(a) Have a clear starting point. (b) Be careful not to overcount/undercount.
Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but
wants to avoid putting any pair of twins into the same team. Subject to this condition:
i) In how many ways can he split them into two teams of six?
ii) In how many ways can he split them into three teams of four?
i) Starting Point:
Further manipulation
to avoid overcounting:
Team 1 must contain one of each of the twins (since we
can’t have both of any pair of twins). For each of the twins
we have 2 choices, so that’s 26 ways of picking.
The teams are indistinguishable (i.e. we don’t have have
team names ‘Team 1’ and ‘Team 2’ – if all of Team 1 moved
to Team 2 and vice versa, it would be considered the same
possibility). So each 2 possibilities only counts as 1.
That gives 26 / 2 = 25 = 32 possibilities.
Tip #6: Multi-step combinatoric problems
For arrangement problems involving multiple steps:
(a) Have a clear starting point. (b) Be careful not to overcount/undercount.
Question: How many ways 2n people can be paired off
to form n teams of 2.
Answer:
(2n)!
?n
n! 2
BMO
Round 2
Round 1
Tip #6: Multi-step combinatoric problems
For arrangement problems involving multiple steps:
(a) Have a clear starting point. (b) Be careful not to overcount/undercount.
Question: How many ways 2n people can be paired off to form n teams of 2.
Starting Point:
Of the 2n people, pick n of them to form the first person in
each team. There’s 2nCn ways of picking them.
Further manipulation
to avoid overcounting:
As a second step, we have n! ways of allocating the
remaining people to these teams.
But we’ve overcounted, because in each team, it doesn’t
matter which person was picked first, so for each team
there’s 2 equivalent arrangements. Since there’s n teams, we
must divide by 2n. This gives us:
2nC
(2n)!
n
=
n
n! 2
n! 2n
Note: If we initially numbered the teams 1 to n as if they were distinguishable and in a line, there would be 2nPn ways of putting n of
the 2n people into these slots, then again n! ways of allocating the remaining people to these slots. But then we’d have to both
divide by 2n (because of the duplication described above), but also by n!, since the teams are in fact not distinguishable, and any way
we’d relabel the team names would be equivalent. This gives the same result as above.
Topic 4 – Combinatorics
Part 3: Recurrence Relations
Here, we explore problems that can be defined in terms of another
related problem (usually a smaller problem).
Introduction
A frog has 5 lily pads in front of him in a line. He needs to get to the final lily
pad, and can do so by either by ‘hopping’ to the next lily pad, or ‘skipping’ to a
lily pad 2 ahead. How many ways are there for the frog to get to the final lily
pad?
There are 3 fundamental steps to solving such problems. The first is as such:
1. Identifying the State
and Actions
What variables can we use to define the current
situation? (known as the current ‘state’)
What actions are available in a given state, and
what states do each of them they lead to?
#1 – Identifying the State and Actions
1. Identifying the State
and Actions
State:
Actions:
What variables can we use to define the current
situation? (known as the current ‘state’)
What actions are available in a given state, and
what states do each of these options lead to?
The number of lily pads in front of the frog. Let’s represent this
using a variable n.
2 possible actions:
a. Hop: We’re now in a state where there’s n-1 lily pads remaining.
b. Skip: We’re now in a state where there’s n-2 lily pads remaining.
#1 – Identifying the State and Actions
State:
Actions:
The number of lily pads in front of the frog. Let’s represent this
using a variable n.
2 possible actions:
a. Hop: We’re now in a state where there’s n-1 lily pads remaining.
b. Skip: We’re now in a state where there’s n-2 lily pads remaining.
It’s possible to represent these transitions between states as a ‘state diagram’:
Pads left
= n-1
Pads left
=n
Pads left
= n-2
Pads left
=0
This states/action graph is known
formally as a Deterministic Finite
Automaton. The double border of
the state on the right indicates it’s
a ‘final state’ where we can stop.
#2 – Forming a Recurrence
Using our knowledge of how different actions affect the state, it’s possible to
define some combinatorial property of the problem in terms of another
(typically smaller) problem.
Let’s define a function F which defines the number of hops based on the
current state.
F(0) = 1?
F(1) = 1?
These are known as the base cases (i.e. considering the smallest version of the
problem, in this case when we have 0 or 1 lily pads in front). It might seem odd
that there’s 1 way to get to the final lily pad when we’re already at the final lily
pad! But we’ll see in a second that this makes things work out.
We need base cases when one or more of the actions are no longer available,
e.g. when there’s 1 lily pad left, we can no longer skip.
#2 – Forming a Recurrence
Now we form a recursive relation, where we define F(n) in terms of smaller problem(s).
Forming a recurrence is usually a case of adding together the number of ways
associated with each of the subsequent states, i.e.
F(current state) = F(possible next states)
F(n) = F(n-1) ?+ F(n-2)
If we’d hopped, we’d be in a state where there’s n-1 lily pads, with F(n-1) ways of
getting to the end (by definition of F).
If we’d skipped, similarly we’d have F(n-2) ways of getting to the end.
Since we’re considering both the possibilities that we start with a skip and the
possibilities where we start with a hop, we add these possibilities together.
#3 – Dynamic Programming
We’ve now got a recurrence relation (with base case) from which we can work out the
number of ways of getting to the last lily pad:
F(0) = 1
F(1) = 1
F(n) = F(n-1) + F(n-2)
We need F(5). Using our recurrence:
F(5) = F(4) + F(3)
= F(3) + F(2) + F(2) + F(1)
= F(2) + F(1) + F(1) + F(0) + F(1) + F(0) + F(1)
= F(1) + F(0) + F(1) + F(1) + F(0) + F(1) + F(0) + F(1)
=1+1+1+1+1+1+1+1
=8
This method is computing the result is quite wasteful,
because we’re having to work out F(2) multiple times
for example. Instead of starting from F(5) and using
our recurrence to work down, perhaps we can start
from the base case(s) and work our way up?
#3 – Dynamic Programming
We’ve now got a recurrence relation (with base case) from which we can work out the
number of ways of getting to the last lily pad:
F(0) = 1
F(1) = 1
F(n) = F(n-1) + F(n-2)
F(0) = 1
F(1) = 1
F(2) = F(0) + F(1) = 2
F(3) = F(2) + F(1) = 3
F(4) = F(3) + F(2) = 5
F(5) = F(4) + F(3) = 8
By working from the bottom-up, each F(n)
only had to be calculated once. This
technique is known in the computing biz as
dynamic programming.
Note that it’s sometimes (but not always) possible to form a position-to-term formula. The Fibonacci
Sequence (which is what we found for this problem!) has a formula involving the golden ratio.
Summary
1. Identify the State and
Actions
State: Number of lily pads left.
Actions available: Hop (where in new state, we have 1
lily pad less) or skip (where we have 2 lily pads less)
2. Forming a recurrence
Define a function F(state)
F(current state) = F(possible next states)
Base cases: F(1) = 1, F(0) = 1
Recurrence: F(n) = F(n-1) + F(n-2)
3. Computing the desired
value.
For problems of size 0 (e.g. no
lily pads left), the number of
ways tends to be 1.
Find F(state described in problem)
Work out the value for problems of ascending size
(i.e. working from the ‘bottom up’). If the state has
two variables, form a table.
F(0) = 1, F(1) = 1, F(2) = 2, F(3) = 3, F(4) = 5, F(5) = 8.
BMO Question
Isaac attempts all six questions on an Olympiad paper in
order. Each question is marked on a scale from 0 to 10.
He never scores more in a later question than in any
earlier question. How many different possible sequences
of six marks can he achieve?
State:
Actions:
Consider the state just before you answer a question:
n – the maximum mark for the remaining questions.
k – the number of questions left
?
•For the next question, score the maximum of n marks.
We’re left in a state where we have k-1 questions left,
and the maximum mark is still n.
•Or for the next question, score n-1 marks. We’re left in a
state where we have k-1 questions left, and the
maximum mark is now k-1 (because we’re not allowed to
score higher than the previous question).
•...and so on...
•Or for the next question, we score 0 marks. For the
remaining k-1 questions, the maximum mark is now 0 for
the remaining questions.
?
BMO
Round 2
Round 1
BMO Question
Isaac attempts all six questions on an Olympiad paper in
order. Each question is marked on a scale from 0 to 10.
He never scores more in a later question than in any
earlier question. How many different possible sequences
of six marks can he achieve?
Step 2 is to convert this information into a recurrence relation.
F(n,k) =
F(n, k-1)
+ F(n-1, k-1)
+ ...
+ F(0, k-1)
We include all the possibilities where we lose
no marks on the next question. Then we’re
interested in the number of ways of scoring up
to n marks for the remaining k-1 questions.
We also include the possibilities where we get
one less than the maximum mark on the next
question. Then for the remaining k-1 questions,
we’re not allowed to exceed n-1 marks.
BMO
And so on...
Round 2
Round 1
BMO Question
Isaac attempts all six questions on an Olympiad paper in
order. Each question is marked on a scale from 0 to 10.
He never scores more in a later question than in any
earlier question. How many different possible sequences
of six marks can he achieve?
Now deal with base cases...
F(n,0) = 1
When there’s 0 questions left, we say
there’s 1 way of doing this (as we usually
say for ‘empty’ problems).
F(0,k) =
If the maximum mark is 0, there’s only 1
possible set of marks we can get: a string
of 0s!
1
?
?
(If you’re uncomfortable defining a base case when there’s no
questions, you could have instead used F(n,1) = n+1 since for
the 1 question we could have scored between 0 and n marks,
i.e. n+1 different marks)
BMO
Round 2
Round 1
BMO Question
Now we need to compute F(10,6) since there’s a maximum of 10
marks and 6 questions.
Since there’s two variables here instead of one, let’s form a table:
F
k=0
1
2
3
4
5
6
F(2,2) for example is
calculated using
F(0,1)+F(1,1)+F(2,1).
But notice that because
F(1,2) = F(0,1)+F(1,1), we
can calculate F(2,2) using
F(2,1)+F(1,2), i.e. adding
the values above and to
the left.
n=0
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
2
1
3
6
10
15
21
28
3
1
4
10
20
35
56
84
4
1
5
15
35
70
126
210
5
1
6
21
56
126
252
462
6
1
7
28
84
210
462
924
7
1
8
36
120
330
792
1716
8
1
9
45
165
495
1287
3003
BMO
9
1
10
55
220
715
2002
5005
Round 2
10
1
11
66
286
1001
3003
8008
Round 1
These are the
triangular numbers!
These are the
These are the
tetrahedric numbers! pentatope numbers!
We can therefore rewrite
the recurrence as:
F(n,k) = F(n-1,k)+F(n,k-1).
(Interesting note – if you tilt your
head left, notice you get Pascal’s
Triangle! Therefore 16C6 would
quickly yield the correct answer
of 8008)
BMO Question
Question
10
9
8
7
Score
6
5
4
0
1
2
3
4
5
6
But there’s a ridiculously
elegant alternative that uses a
more combinatorial approach.
Imagine a path through a
network like on the left, where
only right and down
movements are allowed.
What question scores does this
path represent?
8, 7, 7,?3, 1, 1
3
And how many such paths are
there?
2
(Hint: think of each path as a sequence
of moves. What’s the length of all such
sequences? What are we arranging?)
1
0
16
16
?

6
10
Topic 4 – Combinatorics
Part 4: Compositions and Partitions
Some problems concern how many ways we can divide something
up into parts. For example, how many ways can we express the
number 4 as the sum of positive integers? (e.g. 3 + 1, 1 + 2 + 1, ...)
Composition
A composition of an integer n is the number of ways of expressing it as a
sum of (strictly) positive integers.
Compositions of 5:
5
4+1
3+2
3+1+1
2+3
2+2+1
2+1+2
2+1+1+1
1+4
1+3+1
1+2+2
1+2+1+1
1+1+3
1+1+2+1
1+1+1+2
1+1+1+1+1
The order of the numbers
matters, so 4+1 is a distinct
composition from 1+4.
Notice that we’ve been systematic in writing out the possibilities to ensure we
don’t accidentally miss any (by dealing with an increasingly small first number).
Partition
In contrast, with a partition, the order of the numbers doesn’t matter.
Partitions of 5:
5
4+1
3+2
3+1+1
2+2+1
2+1+1+1
1+1+1+1+1
The order of the numbers
matters, so 4+1 is a distinct
composition from 1+4.
There’s no ‘closed form’
expression that will give the
number of partitions for a
number n. This means there’s no
expression in terms of common
mathematical operations and
functions: addition,
multiplication, exponentiation,
log, sin, etc.
We’ve again been systematic in listing these! One of your homework questions will
be to try and form a recurrence relation for the number of partitions.
Composition
It’s possible to give an expression for the number of compositions of n. This
technique can be used for related problems which we’ll explore.
If we’re considering the compositions of 5, which out five 1s
separated by spaces:
1  1 1 1 1
Fill each box with either a comma (,) or a plus (+). The comma
represents starting a new number in the composition. Then:
,
,
+
+ 1
1  1 1 1
would represent 2, 1, 1, 2, i.e. 5 = 2+1+1+2. This technique covers all
possible compositions.
There’s n-1 gaps (or ‘slots’), so there’s 2n-1 compositions of n.
Composition
We can use a similar technique to represent the ‘wall boundaries’ between
boxes and related problems.
Question: Suppose that n balls are placed at random into n
boxes. Find the probability that there is exactly one empty box.
Big Hint: This above diagram could be represented as O|OO||O, where we
have n “O” symbols and n-1 “|” wall boundary symbols.
?
n(n-1)
Answer = 2n-1
?Cn
Number of box assignments in
which exactly one box is empty.
Number of possible
assignments to boxes.
Composition
We can use a similar technique to represent the ‘wall boundaries’ between
boxes and related problems.
n(n-1)
Answer = 2n-1
Cn
If there’s n boxes and n balls and only one box
is empty, then all other boxes have 1 ball
except for one which has 2. There’s n boxes the
empty one could be, and n-1 remaining boxes
the one with 2 balls could be.
The above arrangement could be represented as O|OO||O by representing the box
boundaries using the symbol |. Then it’s just the case of finding the number of ways of
arranging n ‘O’ symbols and n-1 ‘|’ symbol. In the n+(n-1) total symbol slots, we have n
positions to choose for the ‘O’s.
Combinations with Repetition
Suppose you want to choose  objects where you have  different types of
objects to choose from, where you have an unlimited supply of each type.
e.g. You want to buy 4 tins of drink. There’s 3 different types of drink you
can buy (Fanta, Sprite, Coke). The shop has unlimited stock of each.
Possible selection
Possible selection
Possible selection
Combinations with Repetition
We could model this using boxes and balls.
The boxes would represent  each type?of drink
The balls would represent  a choice of?drink
OO|O|O
So combinations for 
objects and  types:
+−
?

Rings Revisited!
Ring problems involving indistinguishable objects tend to be quite
cumbersome, since there’s often no ‘closed form’ formula for the solution.
Question: How many ways are there of arrangement 4 red beads and 4
yellow beads on a bracelet*.
Answer:
8?
Similarly to the sisters problem, we place the
beads of one colour on the bracelet first, then
consider how we can add the other beads in
between.
If we start with the 4 red beads then we can fit
the yellow beads into the gaps, either 3 all in
one gap, 2 is one gap and 1 in another, or 1 in
each of the four gaps.
* Convention is that ‘bracelet’ implies that the arrangement is considered the same when we flip the bracelet
over, whereas a necklace tends to be one way up, so the arrangements would be considered distinct.
Rings Revisited!
Ring problems involving indistinguishable objects tend to be quite
cumbersome, since there’s often no ‘closed form’ formula for the solution.
Question: How many ways are there of arrangement 4 red beads and 4
yellow beads on a bracelet.
Let (a,b,c,d) represent the number of yellow
beads in each of the four gaps. Then distinct
arrangements (when we consider equivalent
ones due to rotation/reflection) are:
(4, 0, 0, 0)
(3, 1, 0, 0)
(3, 0, 1, 0)
(2, 2, 0, 0)
(2, 0, 2, 0)
(2, 1, 1, 0)
(2, 1, 0, 1)
(1, 1, 1, 1)
This corresponds
to the diagram on
the left
Note: The possible (a,b,c,d) look a bit like the partition of the number 4 (subject to some restrictions),
which is why having a general closed-form formula for n beads is not possible. Damn!
Topic 4 – Combinatorics
Epilogue: Geometric Arrangements
Finally, we look at problems involving shapes, and other visual
structures such as trees.
Counting Internal Shapes
When asked to find smaller shapes within a larger shape, make sure your
counting is structured, in order to identify a pattern.
Classic Question: How many squares
(of any size) are in this diagram?
Answer:
?
30
More generally, what about an n x n grid?
Answer:
1
n(n+1)(2n+1)
?
6
1 x 1 squares: n2
2x2 squares: (n-1)2
...
n x n squares: 1
This is the sum of the first n square numbers, which is a standard result that you
should absolutely remember!
Counting Internal Shapes
Sometimes, you get different sequences for odd and even n.
How many triangles?
Counting Internal Shapes
Up triangles
Down triangles
n=1
1
-
n=2
3+1=5
1
n=3
6 + 3 + 1 = 10
3
n=4
10 + 6 + 3 + 1 = 20
6+1=7
15 + 10 + 6 + 3 + 1
= 35
10 + 3 = 13
21 + 15 + 10 + 6 + 3 + 1
= 56
15 + 6 + 1 = 22
n=5
n=6
Notice that a new size
of triangle appears
every other figure. This
suggests we’ll need
separate formulae for
odd and even n.
Counting Internal Shapes
Since these numbers involve the sum of triangular numbers, and the formula for
triangular numbers is quadratic, it suggests that our overall formula will be cubic.
We can use the numbers we have so far to work out the coefficients (we have 4
unknowns, so require 4 of the numbers in the sequence).
n
2
F(n) 6
4
6
8
n
27
78
170
F(n) 1
F(n) = ax3 + bx2 + cx + d
6 = 8a + 4b + 2c + d
27 = 64a + 16b + 4c + d
...
Solving these simultaneous
equations gives us a, b, c and d.
This gives F(n) = 1 n(n+2)(2n+1)
8
1
3
5
7
13
48
118
1=a+b+c+d
13 = 27a + 9b + 3c + d
...
This gives F(n) = 1 [n(n+2)(2n+1) – 1]
8
The Catalan Numbers
The Catalan Numbers are useful in solving a number of combinatorial
problems involving both shapes and trees.
Imagine a string of n X’s and n Y’s (giving a word of length 2n),
such that for any prefix of the word, the Y’s never outnumber
the X’s.
XXYXYY
A prefix of a word is any sequence of letters from
the start of the word. Here the Y’s don’t
outnumber the X’s.
Then the nth ‘Catalan number’ is the number of such possible
words of length 2n.
The Catalan Numbers
The Catalan Numbers are useful in solving a number of combinatorial
problems involving both shapes and trees.
The nth Catalan number is given by:
_1_ 2n
Cn =
n+1 n
( )
First few Catalan numbers:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796,
58786, 208012, ...
The Catalan Numbers
XXYXYY
(()())
We could replace the X’s by opening brackets and Y’s by closing brackets. Thus,
the Catalan numbers also give us the number of possible valid bracketings.
Other applications:
The number of binary trees with
n+1 leaves (binary meaning that
each parent has 2 children):
Problem 4 on Worksheet 3 (involving
non-overlapping handshakes across a
table) also yields the Catalan numbers!
The number of ways a convex
polygon with n+2 sides can be cut
into triangles:

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