Topic 14: Pigeonhole Principle
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 18th July 2013
Slide guidance
Key to question types:
SMC
Senior Maths Challenge
Uni
Questions used in university
interviews (possibly Oxbridge).
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.
Classic
Maths Aptitude Test
STEP
Admissions test for those
applying for Maths and/or
Computer Science at Oxford
University.
University Interview
Classic
Well known problems in maths.
STEP Exam
Exam used as a condition for
offers to universities such as
Cambridge and Bath.
Slide guidance
?

Any box with a ? can be clicked to reveal the answer (this
works particularly well with interactive whiteboards!).
Make sure you’re viewing the slides in slideshow mode.
For multiple choice questions (e.g. SMC), click your choice to
reveal the answer (try below!)
Question: The capital of Spain is:
A: London

B: Paris

C: Madrid

Starter Question
Show that if I pack 5 different numbers from 1 to 8,
at least two of them add up to 9.
1 2 3 4 5 6 7 8
Solution: There are four pairs of numbers which add up to 9:
{1, 8}, {2,7}, {3,6}, {4,5}.
For the numbers we choose, each must belong to
one of the four pairs. But there’s 5 numbers, so two
must belong to the same pair.
?
The Pigeonhole Principle
Imagine that we have a number of slots, or ‘pigeon holes’
in which we can put letters.
J Frost
A Dumbledore
A Filch
S Snape
R Hagrid
R Lupin
F Flitwick
P Sprout
The pigeonhole principle is very simple: if we have more letters than
pigeonholes, then one of the pigeonholes must have at least one letter.
* The Pigeonhole Principle is also sometimes known as Dirichlet’s Box Principle.
The Pigeonhole Principle
There’s many problems in which we can model as
pigeonholes. If we take the previous problem:
8
{1,8}
Each pigeonhole
represents a
number pair.
2 7
{2,7}
6
{3,6}
The letters
represent the
numbers from 1
to 8 we choose.
4
{4,5}
By the pigeonhole
principle, as we used
more numbers then
pairs, two of the
numbers must be in
the same pair.
In general, we use the pigeonholes to model ‘possible values’, and letters to
model ‘items’, which is put into a pigeonhole hole according to its value.
Another Example of Modelling
Question: Show that if we pick 5 cards from a pack, two
must be of the same suit.
Club
Each pigeonhole
represents a suit.
Diamond
Heart
The letters represent
the cards we choose.
We put it in the correct
pigeonhole according to
its suit.
Spade
Since we have more
cards than suits, one
pigeonhole must
have at least two
cards in it, and thus
at least two cards of
the same suit.
Dijkstra’s Principle
Given there are 8 pigeonholes and 18 letters, what can we
say about the maximum number of letters in a pigeonhole?
J Frost
A Dumbledore
A Filch
S Snape
R Hagrid
R Lupin
F Flitwick
P Sprout
When we spread out the letters as much as possible, we found that the maximum
number of letters was 3.
Dijkstra’s Principle (just a variant of the ‘Pigeonhole Principle’) states that all the
maximum number of letters must be at least the average number of letters.
Dijkstra’s Principle
J Frost
A Dumbledore
A Filch
S Snape
R Hagrid
R Lupin
F Flitwick
P Sprout
There average number of letters in each pigeonhole is 18/8 = 2.25. But
since the maximum number of letters must be at least this, the minimum
this maximum can be is the nearest whole number above, i.e. 3.
In general, if there are n letters and k pigeonholes, there maximum
letters in a pigeonhole must be at least:
These special brackets are
known as the ‘ceiling’ function,
which rounds the number up.
Dijkstra’s Principle
A simple example:
Question: There are 1000 people in the room. What is the minimum
number of people that share the same birthday?
Solution: 3. By the pigeonhole principle, the maximum number of
birthdays is at least 1000/365 = 2.74.
? Therefore at least 3 people share
the same birthday.
Restricted Allocation
Sometimes how we put letters in some pigeonholes restricts how we put
them in others.
There are n people at a party, where each person can shake hands with
others at the party (or if they’re antisocial, no one at all), although they
can only shake a given person’s hand at most once. Show that at least 2
people must have had the same number of handshakes.
Jim
0
Bob
1
....
Tim
n-1
n
Solution: Each pigeonhole represents the quantity of handshakes a person had.
Consider the ‘edge cases’: if someone shook hands with no one, the no one can
? Click
have shook hands with everyone,
andfor
vicesolution
versa. Thus?we only have n-1 available
pigeonholes, but n items, so one pigeonhole must have at least 2 items.
Source: Uni + Classic
Minimal Arrangements
Sometimes we’re required to do some calculation (e.g. the sum) involving
letters in the minimal valid pigeonhole arrangement.
20 cities are visited by up to 20 people each, such that no two cities are
visited by the same number of people.
a) Minimum total number of visitors to the cities:
?
190
b) Maximum total number of visitors to the cities:
?
210
The Lowest Maximum
When we want to consider the lowest maximum, we spread out the letters
as evenly as possible over the pigeon holes.
Question: A cinema sells 31 tickets in total one evening, and each movie
can be watched by up to 3 people (but won’t screen if no one shows up).
Show that the minimum number of movies seen by the same number of
people is 6.
Let’s deconstruct the problem:
What are the pidgeonholes?
The number of people?attending a screening.
What are the letters?
Each movie/screening.?
Source: Frosty Special
The Lowest Maximum
When we want to consider the lowest maximum, we spread out the letters
as evenly as possible over the pigeon holes.
Question: A cinema sells 31 tickets in total one evening, and each movie
can be watched by up to 3 people (but won’t screen if no one shows up).
Show that the minimum number of movies seen by the same number of
people is 6.
This means that
Bat
man
1
Spider
man
2
Meerkat
man
Meerkatman was
seen by 3 people.
This letter thus
represents 3 tickets.
3
The arrangement that will minimise the maximum number of letters in a
pidgeonhole (i.e. minimising the number of movies same by the same number of
people) involves putting 1 letter in each pidgeonhole each time.
Each time we do so, this represents 1 + 2 + 3 = 6 tickets.
Source: Frosty Special
The Lowest Maximum
When we want to consider the lowest maximum, we spread out the letters
as evenly as possible over the pigeon holes.
Question: A cinema sells 31 tickets in total one evening, and each movie
can be watched by up to 3 people (but won’t screen if no one shows up).
Show that the minimum number of movies seen by the same number of
people is 6.
Bat
man
Cat
man
Dog
man
1
Fish
man
Rat
man
Spider
Tiger
Vole
man
man
man
Fox
Lima
man
man
Rabbit
Lice
Ant
man
man
man
Ox
Ham
man
man
2
3
Given that putting one letter in each pigeonhole represents 6 total tickets, we can put
Toucan
5 letters in each ticket to get us to 30 tickets.
man
We have 1 ticket left. This must have been for a movie watched by 1 person. And thus, six movies were
watched by the same number of people. (Although note that there’s other solutions in which 6 movies
were watched by the same number of people, e.g. Once we got to 4 letters in each hole, we could have
put 2 letters in the ‘3 hole’ and one in the ‘1 hole’ – the point is that we’ve proved that 6 is the minimum)
BMO Problem
Question: Let S be a subset of the set of numbers {1, 2, 3, …
, 2008} which consists of 756 distinct numbers. Show that
there are two distinct elements, a, b of S such that a + b is
divisible by 8.
Things that might be going through your head when you see this
problem:
“What are my pigeonholes and what are my letters?”
“How might I be restricted in putting letters into pigeonholes?”
“It sounds like from the question that we can have 755 numbers
without any two of them adding to give a multiple of 8. The
problem is therefore equivalent to ‘show that 755 is the
maximum numbers of numbers we can choose without this
happening’”.
BMO
Round 2
Round 1
BMO Problem
Question: Let S be a subset of the set of numbers {1, 2, 3, … , 2008} which
consists of 756 distinct numbers. Show that there are two distinct elements, a, b
of S such that a + b is divisible by 8.
10
80
26
0
1
2
3
4
5
6
7
Whether the sum of a and b is divisible by 8 is equivalent to using the remainder of
each of the numbers when we divide each by 8. e.g. If 15 + 9 is divisible by 8, then so
is 7 + 1 (we’ll encounter modulo arithmetic in the ‘Number Theory’ topic).
So make the pigeonholes 0-7, and put a number in this hole if it’s remainder is that
number when dividing by 8.
BMO Problem
Question: Let S be a subset of the set of numbers {1, 2, 3, … , 2008} which
consists of 756 distinct numbers. Show that there are two distinct elements, a, b
of S such that a + b is divisible by 8.
10
80
0
4
We’re restricted in how we
can put letters in holes.
26
1
5
2
6
3
7
If there’s any in the 3 hole, we
can’t put any the 5 hole because
then we’d have two numbers
which add to 8.
Similarly, we could only have one
number in the 4 hole; if there
were two they’d again add up to
8. The same applies for the 0
hole.
BMO Problem
Question: Let S be a subset of the set of numbers {1, 2, 3, … , 2008} which
consists of 756 distinct numbers. Show that there are two distinct elements, a, b
of S such that a + b is divisible by 8.
10
80
0
4
26
1
5
2
6
3
7
There’s 2008 numbers, so 2008/8 =
251 letters belong to each
pigeonhole (e.g. all the multiples of
8, of which there are 251, would go
in the ‘0’ pigeonhole).
We would put 251 letters in either
hole 1 or 7. Similarly, we could put
251 in hole 2 or 6. And another 251 in
hole 3 or 5. And we could put 1 in
holes 0 and 4. That gives us 251 + 251
+ 251 + 1 + 1 = 755. This therefore is
the maximum number before two of
the numbers add to a multiple of 8.
With 756, we’d therefore violate this
property.
BMO Problem
Question: A booking office at a railway station sells tickets
to 200 destinations. One day, tickets were issued to 3800
passengers. Show that
(i) there are (at least) 6 destinations at which the passenger
arrival numbers are the same;
(ii) the statement in (i) becomes false if ‘6’ is replaced by ‘7’.
Things that might be going through your head when you see this
problem:
“What are my pigeonholes and what are my letters?”
“What’s the average number of people at each station?”
“A station is limited in the number of people that can buy a
ticket for there. For example, if 3800 all bought a ticket for the
same station, there wouldn’t be 200 destinations! So we must
have to restrict the allocation of letters to pigeonholes in some
way.”
BMO
Round 2
Round 1
BMO Problem
Question: A booking office at a railway station sells tickets to 200 destinations. One day, tickets were
issued to 3800 passengers. Show that (i) there are (at least) 6 destinations at which the passenger
arrival numbers are the same; (ii) the statement in (i) becomes false if ‘6’ is replaced by ‘7’.
Kings
Paris
Cross
0
Bath
1
....
Oslo
Hull
18
19
20
...
My solution: Model the pigeonholes as the number of tickets for a station. Let the
letters be the stations (so Bath goes in pigeonhole 1 if 1 person is going to Bath).
If there’s 3800 passengers, and 200 destinations, there’s an average of 19 people per
station. Note that this relates to the average value of the pigeonhole, NOT the average
number of letters in each pigeonhole.
The key to this strategy is thinking how we allocate letters to pigeonholes so the
average pigeonhole value is 19.
BMO Problem
Question: A booking office at a railway station sells tickets to 200 destinations. One day, tickets were
issued to 3800 passengers. Show that (i) there are (at least) 6 destinations at which the passenger
arrival numbers are the same; (ii) the statement in (i) becomes false if ‘6’ is replaced by ‘7’.
Kings
Cross
0
Bath
1
....
Oslo
Hull
Kings
Cross
37
38
39
...
We want to minimise the number of letters in each pigeonhole. The way to do this is
put one letter in each of the pigeonholes 0 to 38. As highlighted earlier, this ensures
the average people per destination is still 19 whilst minimising the number of stations
with the same number of passengers.
BMO Problem
Question: A booking office at a railway station sells tickets to 200 destinations. One day, tickets were
issued to 3800 passengers. Show that (i) there are (at least) 6 destinations at which the passenger
arrival numbers are the same; (ii) the statement in (i) becomes false if ‘6’ is replaced by ‘7’.
Kings
Paris
Cross
Bath
Madrid
0
1
....
Oslo
Berne
37
Hull
Hell
38
Kings
Cross
39
...
We keep adding in this way for each 39 stations.
Each time we do, this represents 0 + 1 + 2 + 3 + ... + 38 tickets. Using the standard
summation formula, this gives 741 tickets. 741 goes into 3800 five whole times, so we
have 5 letters in each of the pigeonholes 0 to 38 and have 741x5 = 3705 tickets.
How do we deal with the remaining 95 tickets? The average per station needs to be 19,
which means we’re spreading these over 95/19 = 5 stations. So using the 17, 18, 19, 20
and 21 pigeonholes will do to keep the quantities of tickets for each station spread out..
Thus in the minimal case, 5 quantities of tickets will have 6 stations with that quantity.
This proves (ii) as well, because we’ve proved this minimum can be 6.
One final problem…
The mathematician Paul Erdős once asked a
student what he wanted to when he graduated.
The student said he wanted to be a mathematical
researcher. Erdős said in response:
Let 𝑛 ∈ ℕ. Prove that if you take a
set 𝐷 of 𝑛 + 1 numbers from
{1,2, … , 2𝑛} then you can find
distinct 𝑖, 𝑗 ∈ 𝐷 such that 𝑖 | 𝑗.
( 𝑖|𝑗 means that 𝑖 divides 𝑗)
i.e. Show that if you pick 𝑛 + 1 numbers from 1 to 2𝑛, one
must divide another.
Big Hint: Suppose that each pigeonhole contains the integers of the
form 𝑘2𝑞 , for each odd number 𝑘, i.e. the first contains numbers of
the form 1 × 2𝑖 , the second of the form 3 × 2𝑖 and so on.
Suppose n is 5. Then we have 5 odd numbers, 1, 3, 5, 7,
9, and the numbers in each pigeonhole (using the hint
above) will be:
1, 2, 4, 8
5,10
?
3, 6
7
{9}
We have 5 pigeonholes (as there are 5 odd numbers), but
5+1 = 6 numbers to pick. So by the pigeonhole principle,
we’ll have chosen at least 2 numbers from one set. And
clearly the numbers in each set divide each other, since
𝑐2𝑖 |𝑐2𝑗 when 𝑖 ≤ 𝑗.
Summary
For some problems, we use can use pigeonholes to represent different
values, and letters to represent items can have one of these values.
Be careful how you model these.
There are two forms of the pigeonhole principle:
1. If there are more items than pigeonholes, at least one of the
pigeonholes has more than one item in it.
(Also known as ‘Dirichlet’s Box Principle’)
2.
If there are n items and k pigeonholes, then the maximum number
of items in a pigeonhole is at least the average, n/k.
(This is Dijkstra’s take on the pigeonhole principle, although this variant is also know as the ‘Fubini Principle’)
Often, putting items in one pigeonhole prevents items from being
added to another.
When considering problems in which we wish the minimise the
maximum number of letters in a hole, spread the letters out as much
as possible, without violating any given/implied constraints.
Recommended Resource
http://www.cut-the-knot.org/do_you_know/pigeon.shtml has a gigantic
collection of problems from past international Olympiads (and
elsewhere) concerning the pigeonhole principle.