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Math 216A Notes, Week 1
Disclaimer: These notes are not nearly as polished (and quite possibly not nearly as correct) as a published
paper. Please use them at your own risk.
1. Introduction
In a sense much of what we will be doing this term falls under (indirect) counting. Some of the applications
of this are clear – if you’re running a computation you want to know in advance how long the calculation
will take. But there’s also more subtle uses.
Suppose you were trying to build an (expensive) communications network connecting n nodes, or vertices.
Here’s an example of exactly the sort of design you don’t want to use:
Here S and T both have size n/2, and every vertex in S connects to every other vertex in S (and likewise
for T ; we’ll sometimes refer to this as meaning that S and T both are complete graphs on n/2 vertices),
however there is only one edge connecting S and T . This has two problems.
• We’re constructing a huge number of edges that may not be really necessary
• If the one edge between S and T fails, the graph disconnects into two large components
It’s possible to fix each of those problems individually (e.g. turning the graph to a single path fixes the first,
and turning it into a complete graph on n vertices fixes the second), but can we fix both of them at once?
Theorem 1. (Pinsker, 1973 and Barzdin/Kolmogorov, 1967) If n is large enough and even, there are
3−regular graphs on n vertices (graphs where every vertex has exactly 3 neighbors) such that for any subset
A of size at most n/2, at least 0.1|A| edges must be removed to disconnect A from rest of the graph.
Proof. (sketch): The number of 3−regular graphs is much larger than the number of graphs which fail to
satisfy the disconnection property. So in fact a randomly chosen 3-regular graph satisfies this property with
probability close to 1.
Topics to be Covered:
In a sense this is going to be a survey course that jumps around a fair amount from topic to topic. The key
topics I’m planning on covering are
• Enumerative Combinatorics, including both basic counting and sieve methods such as inclusionexclusion
• Probabilistic Methods (both non-constructive proofs like the one above and general methods of
saying ”almost all objects in a class have a certain property”)
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• Ramsey Theory. Here we’ll be considering results along the lines of ”if you color X with finitely
many colors, and X is large enough, some monochromatic structure Y must appear. For example, X
could be the edges of a complete graph, and Y be a (fixed finite) graph. Or X could be the integers
from 1 to n, and Y be an arithmetic progression of (fixed finite) length.
• Additive Combinatorics (combinatorial number theory). Here’s the sort of question we might be
considering then. For any finite set of integers (or residues mod p) A, we can consider the sets
A+A
AA
:= {x + y|x, y ∈ A}
:= {xy|x, y ∈ A}.
If A were a subring, |A+A| and |AA| would both be equal to |A|. But the integers have no (nontrivial)
finite subrings, so in fact both of these sets are larger than A. It’s possible to make |A + A| not too
large (2|A| + 1), by taking A to be an arithmetic progression. But then |AA| is close to |A|2 . It’s
possible to make |AA| not too large (again 2|A| + 1), by taking A to be a geometric progression. But
then |A + A| is close to |A|2 . Can we make both small at the same time?
• Other Topics People Are Interested In
My rough guess is that I’ll be spending about 2-3 weeks on each of these, but I really don’t have a rigid
expectation on what I’ll be covering when and this could change.
Textbook:
There is no official text for this course. I expect the content at various points to parallel material in
-Richard Stanley’s Enumerative Combinatorics, Vol. 1 (available online at http://www-math.mit.edu/∼rstan/ec/ec1/
)
-Noga Alon and Joel Spencer’s The Probabilistic Method (available online through the UCR Library)
-(Possibly) Terence Tao and Van Vu’s Additive Combinatorics
-(Possibly) J. H. Van Lint and R. M. Wilson’s A Course in Combinatorics
There will be a set of course notes maintained by the class (see below):
Course Requirements: There are two main requirements for the course.
1. Two take-home problem sets, one around midterm week and one around finals week.
2. Serving as scribe for one week’s worth of course notes. I will post the first week of notes on the course
website, and you can either use that as a template or write the notes in some other format.
2. A Few Warm-Up Problems
Let’s start with a few warm-up counting problems. Later on we’ll try and find a more general framework to
fit these in.
Warm Up 1: We have k different kinds of postcards, and want to send one postcard to each of n people.
How many ways can this be done?
Solution: There are k choices which card to send to person 1, k choices which card to send to person 2,
and so on. Multiplying gives k n .
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Warm Up 2: Suppose we only have one postcard of each type, so have to send different kinds to each
person. Now how many?
Solution: There are k choices which card to send to person 1, k − 1 choices which to send to person 2, and
so on. Multiplying gives k(k − 1)(k − 2) . . . (k − n + 1). (This is sometimes called (k)n )
Warm Up 3: Suppose that you have k identical postcards that you want to send to n people, where each
person gets at most 1 postcard. How many ways can you choose k people to get postcards?
Solution: Suppose we sent out the postcards in order. There’s n choices for who gets the first, n − 1 for
who gets the second, and so on. So there’s n(n − 1) . . . (n − k + 1) = (n)k ways of sending them out. But for
each set of k people there’s k(k − 1)(k − 2) . . . (1) different orders we could have sent the postcards in. So
once we correct for overcounting there’s
n!
n
n(n − 1) . . . (n − k + 1)
=
:=
k(k − 1)(k − 2) . . . (1)
k!(n − k)!
k
Warm Up 4: Suppose that we no longer require people to receive at most one postcard. Now how many
ways are there?
Solution: We can solve this by bijection, placing the objects in one to one correspondence with something
we already know how to count. To each way of giving the cards out we can associate a sequence of k cards
and n − 1 dividers (Person 1 gets all the cards before the first divider, person 2 all the cards between there
and the second divider, etc.).
(5, 2, 0, 1) ↔ • • • • •| • •||•
So the number of ways is the
same as the number of ways to place n − 1 dividers in a sequence of length
(n − 1) + k, which is n−1+k
n−1 .
Warm Up 5: Still k identical postcards, still n people, but now each person needs to get at least one
postcard. Now how many ways?
Solution: For every way of giving out k postcards in this fashion, there’s a corresponding way of giving
k − n postcards as in warm up 4 – just
away
take
one postcard from each person. This correspondence goes
k−1
both ways, so the total is n−1+(k−n)
=
n−1
n−1 .
3. The General Framework – Stanley’s ”Twelvefold Way”
Many counting problems can be reduced to the following form: Given n and k, count the number of functions
from N = {1, 2, . . . , n} to K = {1, 2, . . . , k} satisfying some list of conditions. Equivalently, how many ways
are there to place n balls in k bins, subject to certain conditions? The answer to this question depends not
only on the conditions you impose, but also on what you really mean by ”how many ways”...when are two
different ways considered equivalent? Here are four natural ways of treating equivalence:
• All the balls are distinguishable, and all the bins are distinguishable. Two placements are the same
only when every ball is in exactly the same bin in each
• The bins are distinguishable, but the balls aren’t. All that matters is how many balls are in each
bin.
• The balls are distinguishable, but the bins aren’t. All that matters is which balls are with each other
ball.
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• The balls and bins are both indistinguishable. All that matters how many bins have 1 ball, how
many have 2...
For example, if you have 3 balls A, B, and C and 2 bins, here’s 4 of the different ways you could arrange
them:
Bin 1
AB
AC
C
B
Bin 2
C
B
AB
AC
In the first case, all four of these ways are different. In the second, ”AB/C” is considered the same as ”AC/B”.
In the third, ”AB/C” is the same as ”C/AB”. In the fourth, all four ways are considered equivalent.
In the warm-up problems, we also saw two natural conditions we could place: The functions could be injective
(at most 1 ball per bin), or surjective (at least 1 ball per bin). Throw in the case where there’s no conditions
at all, and that gives us 12 problems to consider. We solved 5 of them above very quickly. Here’s a slightly
more complicated problem.
4. Stirling Numbers
Problem: How many ways are there to partition a set of size n into exactly k smaller sets, if the order of
the smaller sets don’t matter? In terms of the framework above this corresponds to surjective functions in
a ”balls distinguishable, bins not” situation.
The solution to this problem is a called a Stirling Number of the Second Kind, and is denoted S(n, k). One
way of computing them starts with the following:
Claim: For any n ≥ k ≥ 0,
n
X
S(n, k)x(x − 1) . . . (x − k + 1) = xn
k=1
Proof. Since both sides are polynomials of finite degree, it suffices to show that they are equal on positive
integers x. For a given x, the right side counts the number of functions from {1, . . . , n} to {1, . . . , x}. Each
such function is characterized by
• A given k which is the size of the range of the function
• Which set of k numbers is the image of the function
• Which numbers get mapped to each number in the range
For a given k, there’s xk choices for the range, then S(n, k)k! choices for which numbers get mapped to which
place (S(n, k) choices which numbers are together, then another k! choices where to map which numbers)).
Adding up over all k, the total number of functions is, as desired,
X
X
x
S(n, k)k!
=
S(n, k)x(x − 1) . . . (x − k + 1).
k
k
k
We can invert this into a formula for S(n, k) using a bit of algebraic trickery:
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Theorem 2.
k
1 X
k−j k
(−1)
S(n, k) =
jn
k! j=0
j
Proof. We plug in our previous claim to get that the right hand side of the above expression is
X
j
k
k
1 X
1 X
k−j k
n
k−j k
(−1)
j
=
(−1)
S(n, r)j(j − 1) . . . (j − r + 1)
k! j=0
j
k! j=0
j r=1
=
j
k X
X
(−1)k−j S(n, r)
j=0 r=1
=
1
(k − j)!(j − r)!
k X
k−r
1
(−1)k−j
S(n, r)
(k
−
r)!
k
−
j
r=1
j=r
k
X
Here the inner sum in the first inequality only goes up to j because for r > j the product is 0. If we let
m = k − r and t = k − j, the inner sum can be written as
m
X
t m
(−1)
,
t
t=0
which can be seen to be equal to 0 for all m > 0 by two different ways:
1. Using the binomial theorem
(x + y)m =
m X
m t m−t
xy
t
t=0
with x = 1 and y = −1.
2. Bijectively. The summand (−1)t m
t counts subsets of size t with a sign of +1 is t is even, and −1 if t
is odd. Consider the following operation φ: If 1 ∈ S, then φ(S) = S − {1}. If 1 ∈
/ S then φ(S) = S ∪ {1}.
Since applying φ twice returns you to the same set, φ divides the subsets into 21 m
2 pairs (S, φ(S)). Each
pair contains one set of odd cardinality and one set of even cardinality, so there’s an equal number of sets
contributing +1 and −1.
So the inner sum disappears unless k = r, in which case we’re left with just S(n, k).
While correct, this theorem is a bit unsatisfying in that we seemingly needed to know the right hand side in
advance to derive it. We’ll return later on with a more systematic approach.
An alternative way of computing Stirling numbers is to use recursion. We can get a partition of {1, . . . , n}
into k parts in two different ways:
• Start with a partition of {1, . . . , n − 1} into k − 1 parts, then add {n} as its own part. There’s
S(n − 1, k − 1) ways we can do this
• Start with a partition of {1, . . . , n − 1} into k parts, then add n to one of those k parts. There’s
kS(n − 1, k) ways of doing this (S(n − 1, k) choices for the initial partition, k choices where to add
n)
So we have S(n, k) = S(n − 1, k − 1) + kS(n − 1, k). Starting with the base cases S(n, 1) = S(n, n) = 1, we
can use this to compute S(n, k).
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So far we’ve managed to fill in a fair amount of the table below
Elements of X Elements of K
dist.
dist.
indist.
dist.
dist.
indist.
indist.
indist.
Any f
kn n+k−1
Injective f Surjective f
(k)n
? k
n−1
k−1
n
k−1
?
?
?
?
S(n, k)
?
There’s a few more low-hanging pieces of fruit we can pick off now too.
-If we consider set partitions into k parts, but now the order of the parts matters (row 1, column 3), there’s
exactly k! ways to order each of the S(n, k) partitions, for a total of k!S(n, k).
-If we remove the surjectivity requirement in row 3, we effectively are considering set partitions into at most
k parts. One expression for this is S(n, 1) + S(n, 2) + · · · + S(n, k) (not the simplest to work with, but
sometimes we have to take what we can get).
-If we require injectivity in row 3, then effectively there either is a function (if n ≤ k) or there isn’t (if n > x).
So the answer’s either 1 or 0, depending. Ditto for row 4.
What’s left is the rest of the fourth row.
5. Partitions into a Fixed Number of Parts
Say, for example, we want to place 4 balls in 2 identical bins so that each bin gets at least one ball. There
are two ways of doing it: Either split the balls 2, 2 or split them 3, 1. These correspond to the ”partitions”
4 = 2 + 2 and 4 = 3 + 1.
Let pk (n) denote the number of partitions of n into k parts. By the above, we have p2 (4) = 2. Just like with
Stirling numbers, we can approach pk (n) by recursion by dividing the partitions into two classes. Now our
classes are:
• Partitions which contain a part of size 1. Deleting that part, we’re left with a partition of n − 1 into
k − 1 parts, so there’s pk−1 (n − 1) partitions in this class
• Partitions in which every part have size 2. In this case, we can delete 1 from every part, and we’re
left with a partition of n − k into k parts, so there’s pk (n − k) in this class.
Therefore pk (n) = pk−1 (n − 1) + pk (n − k) (with the initial conditions p1 (n) = 1 and pk (n) = 0 for n < k).
Remark 1. It’s important here that both of the decompositions are bijections.
Just like with Stirling numbers, removing the surjectivity requirement corresponds to taking p1 (n) + p2 (n) +
· · · + pk (n), and our table is complete.
6. More General Partitions, and Generating Functions
Let p(n) denote the number of partitions of n where we place no requirements on the number of parts (so
p(n) = p1 (n) + p2 (n) + · · · + pn (n)). For example, p(4) = 5, since 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1.
And p(200) = 3, 972, 999, 029, 388. To get a handle on the partition function, it’s helpful to consider what’s
termed a generating function
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Definition 1. The Generating Function for a sequence a0 , a1 , a2 , . . . is given by
X
an xn
f (x) =
n≥0
For example, the generating function for the sequence an = 2n is given by
X
1
f (x) =
2n xn =
.
1 − 2x
n≥0
For most purposes, we only will worry about generating functions as what we term ”formal” power series,
meaning that we think of them only as ways of encoding coefficients and don’t care about convergence.
The generating function for p(n) is slightly more complicated.
Theorem 3.
X
p(n)xn
=
n≥0
Y
k≥1
=
1
1 − xk
(1 + x + x2 + x3 + . . . )(1 + x2 + x4 + . . . )(1 + x3 + x6 + . . . ) . . .
Proof. The xn coefficient of the right hand side corresponds to the number of ways of choosing one term
from each sum in the product and multiplying them together to get xn (note that this is actually finite –
once k > n we must choose the 1 from the term corresponding to (1 − xk )−1 to avoid having our exponent
increase beyond n). Suppose that we choose the xkmk summand from the product beginning (1 + xk + . . . ).
Then each way of getting an xn product corresponds to a solution to
n
=
1m1 + 2m2 + 3m3 + . . .
=
(1 + 1 + · · · + 1) + (2 + 2 + · · · + 2) + (3 + 3 + · · · + 3) + . . .
|
{z
} |
{z
} |
{z
}
m1 10 s
m2 20 s
m3 30 s
This in turn corresponds to a partition with m1 parts of size 1, m2 parts of size 2, and so on.
An identical argument can be used to give generating functions for partitions once we stick on various
restrictions. For example, suppose we let pd (n) be the number of partitions of n into distinct parts. So for
example 7 = 6 + 1 = 5 + 2 = 4 + 3 = 4 + 2 + 1, so pd (7) = 5. In terms of generating functions, the restriction
means that we can only have mk = 0 or mk = 1 for each k, and we have
X
Y
pd (n)xn =
(1 + xk ).
n≥0
k≥1
As another example, suppose we let po (n) denote the number of partitions of n where every part is odd. For
example, we can write 7 = 5 + 1 + 1 = 3 + 3 + 1 = 3 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1, so po (7) = 5.
The generating function for po (n) is
X
Y
1
po (n)xn =
.
1
−
xk
k≥1
n≥0
k odd
Notice that po (7) = pd (7). This is not a coincidence!
Theorem 4. For n ≥ 1, we have the number of partitions of n into distinct parts equals the number of
partitions of n into odd parts.
The proof of this theorem gives a good idea of how powerful generating functions can be for problems like
this
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Proof. We have
X
pd (n)xn
=
(1 + x)(1 + x2 )(1 + x3 )(1 + x4 ) . . .
=
Y 1 − x2 1 − x4 1 − x6 1 − x8
...
1 − x 1 − x2 1 − x3 1 − x4
n≥0
k≥1
=
Y
k≥1
=
X
1
1
1
...
1 − x 1 − x3 1 − x5
po (n)xn
n≥0
Since the generating functions are equal, the xn coefficient in the generating function must be equal for each
n.
Exercise: Try and come up with a bijective proof of this result.
An alternating way of viewing generating functions are in terms of Ferrers Diagrams (also sometimes called
Young diagrams). To each partition n = λ1 + · · · + λk , we associate a diagram with λi dots or boxes in row i
(The English tend to put the largest row on the top; the French tend to put it on the bottom). For example,
the partition 10 = 5 + 3 + 1 + 1 corresponds to the diagram
One quick result that can be proven using them is
Theorem 5. The number of partitions of n into exactly k parts is equal to the number of partitions of n
where the largest part equals k.
For example, there are 3 partitions of 6 into exactly 2 parts
and 3 partitions of 6 where the largest part equals 2
Proof. (short version) Turn your head 90 degrees!
Proof. (more detail) To each partition we can associate a conjugate partition gotten by reflecting across the
diagonal (swapping the role of rows and columns). This gives a bijection between the two classes
7. The Pentagonal Number Theorem, and a recursion for p(n)
Suppose we were to change the generating function for pd (n) slightly, instead writing
Y
(1 − xk )
k≥1
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This function now counts the difference between the number of partitions of n into an even number of
distinct parts and the number of partitions into an odd number of distinct parts. The actual evaluation of
this product (in terms of extracting coefficients) was first done by Euler. The proof we give here is due to
Franklin.
The rough idea is to do a pairing argument similar to the one we used for the Stirling Sum (method 2
above). Given a Ferrers diagram, let b be the size of the smallest part, and let s be the size of the 45◦
diagonal starting in the upper right (in the below diagram, b = 4 and s = 3).
Consider a function φ defined as follows:
• If b ≤ s, remove the last row and place it after the last 45 degree diagonal, as in the diagram below.
•
−→
•
•
• • •
• If b > s, reverse the operation, removing the last diagonal and placing it after the last row.
Since φ is its own inverse (check this!), it creates pairs (λ, φ(λ)). Since φ changes the parity of the number
of parts, each pair contributes a net of 0 to the sum (these two properties together are what is referred to
as a ”sign reversing involution”). The trouble is, φ isn’t always well defined, in that applying it as written
can lead to something which isn’t a partition in two cases.
• If b = s and the last diagonal and last row intersect, then once we remove the last row we aren’t left
with a valid Tableaux anymore.
•
−→
•
• • •
•
• If s = b + 1 and the last diagonal and row intersect, moving the diagonal to the bottom leaves us a
partition whose parts aren’t distinct.
•
•
−→
•
• • •
In each of these cases, we have one partition left unpaired. If there are k parts to the partition, the first
situation occurs when n = k + (k + 1) + · · · + (2k − 1) = k(3k−1)
. The latter occurs when
2
k(3k+1)
n = (k + 1) + (k + 2) + · · · + (2k) =
. So what we’ve proven is
2
Theorem 6. (Euler’s Pentagonal Number Theorem)
k(3k−1)
Y
X
k(3k+1)
(1 − xk ) = 1 +
(−1)k x 2 + x 2
k≥1
k≥1
This sort of ”involution” argument is often useful when you’re trying to count things with sign – the hope is
that if you find the right mapping, all the things of one sign will cancel and you’ll be left with a few easily
computable fixed points. More generally, we have
Theorem 7. Suppose we have a function φ on a set S satisfying
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• φ2 (x) = x for all x
• If φ(x) 6= x, then f (φ(x)) = −f (x).
Then
X
x∈S
X
f (x) =
f (x)
x∈S
φ(x)=x
A special case of this is where we’re working over F2 and f (x) = 1. Then this tells us that the parity of |S|
is equal to the parity of the number of fixed points (we’ll see a couple more examples of this next time).
Exercise: Show that the parity of p(n) is the same as that of the number of partitions of n into odd distinct
parts. For example if n = 7 there is only one such partition, and p(7) is odd.
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