Discrete Mathematics II, Combinatorics
Math. 325, Sec. 001, Spring 2018
Homework and Announcements
Handwritten solutions of the midterm
exam and some info about the exam have be posted
at Exam Info here below.
You should print out a copy for your own records.
Solutions on the web.
Just to iterate what I briefly mentioned in class,
if you find a solution of an assigned problem online,
it is perfectly OK to use that as a basis for your own solution.
Just beware of the fact that these are rarely endorsed by
the authors of the textbooks and have most likely not
been reviewed properly. With this in mind the solutions
are many times sketchy and then you need to fill in the blanks
and complete them. (They can even be totally wrong on occasions.!)
-- However, to be kosher you should give credit where credit is due
and indicate clearly if you found the "solution" online. For example:
"The following solution is based on a posted solution I found
at Chegg.com" or something like that.
The Pascal's Pyramid.
To see a clear presentation of what I was writing
on the board regarding the three dimensional Pascal's
pyramid, and how it resmebles the usual Pascal's Triangle,
then goto here.
All handouts for the class will be posted as pdf files
after a bullet sign. All these handouts are copyright
material and are intended for the students of this class only.
IN PARTICULAR, DO NOT POST THESE HANDOUTS ON ANY OTHER
WEBSITE OR INTO AN ARCHIVE DATABASE!
For general information regarding this course, please
read the Syllabus here below (to be posted soon), it should answer all
practical questions, like when tests are, what
the grading policies are etc.
Homework and Solutions
Here below the homework (HW) assignments and selected solutions
will be posted. Many of the problems will be exercises
from our textbook. -- Please be sure to read the solutions.
If you have questions or comments feel free to come
and talk to me.
- Homework for Tuesday, January 30. :
We started discussin some preliminaries and listed four
basic principles related to counting:
(C1) Addition Principle, (C2) Product Principle, (C3) The Negation Principle
and (C4) The Division Principle, whose more general form is the
Pigeon Hole Principle (PHP). We discussed some examples on how
this is used to show the existence of certain scenarios. One
corollary of the general form of the PHP is that if n
pigeons are put into n holes and each hole has at most
one pigeon, then each hole must have a pigeon. This can then be used
to prove the Chinese Reminder Theorem.
- Homework for Tuesday, February 6. :
We continued our discussion of the Pigeon Hole Principle (PHP),
and we discussed a particularly nice application of the PHP to
prove a known theorem by Erdős and Szekeres from 1935.
We then discussed a well-know Theorem of Friends and Strangers
that states the in any group of six or more people, where any two
are either friends or strangers, then one can always find a subgroup
of three people that are all mutual friends or all mutual strangers.
Since this is not always possible in a group of five people, then
we know that six is the smallest number of people we must have so
that we can always find three that are all mutually either friends or
strangers. In the language of Ramsey Numbers this means
have R(3,3) = 6, which in the language of graph theory means
that p = 6 is the smallest number such that for any coloring
of the edges of Kp, the complete graph on p
vertices, with two colors (say red and blue), then we can always find a
red K3 or a blue K3 inside
- Homework for Tuesday, February 13. :
We recalled the number of permutations and the
number of r-permtuations of a set containing
n elements. By a circular permutation we mean
an arrangement of elements in a circular fashion where we ignore
the clockwise and counter clockwise rotation of the circle (that is,
a circular arrangement of the elements and any of its rotations is
counted as one.) When counting such circular arrangement it is often
good to fix a designated element, fix that, and then count the
(linear / usual) permutations of the remaining elements. -- By counting
the number of r-permuations in two ways (1) by directly listing
the r elements from a given set of n elements
on one hand, and (2) by first choosing all the r elements
from the given n and then arrange them in a line on the other
hand, we obtain a formula for the number of subsets containing r
elements of a set of n elements. These are the
binomial coefficients and they will be discussed in more
detail later on in the course.
- Homework for Tuesday, February 20. :
By a multiset we mean a collection of elements where
some elements are identical, or in other words, a collection
of elements where we can have more than one copy of each element.
The notions of a submultiset and when two multisets are
equal works in a way analogus to that of usual sets. The number
of permutations of a given multiset is given by a corresponding
multinomial coefficients, a natural generalization
of the binomial coefficients, since when the number k of types is
equal to 2, then the multinomial coefficients reduce
to the binomial coefficients. A typical problem solved by
multinomial coefficient is to count the number of words (real or not real)
one can make from the letters of "MISSISSIPPI". The multinomial coefficients
also count the number of ways we can partition a given (usual) set
into parts, both in a labeled way or in an unlabeled way if all the
parts have the same cardinality. By an r-permutation
of a multiset S we mean an r-string where
no two elements are listed twice (but elements of the same type can.)
Unfortunately, there is no nice formula/expression for the number
of these (unlike when we are dealing with usual sets.) An
r-combination of a multiset S
is simply a sub-multiset of S that contains r
elements. By counting the number of integer solutions to an equation
we obtain a nice binomial formula for the number of
r-combinations of a given multiset. The formula that we
derived to count the number of integer solutions to an integer
equation is interesting in its own right. -- Recall the definition
of a partially ordered set, or a poset for short. It is a binary
relation on a given set which is reflexive, antisymmetric and transitive.
The notion of a subposet is natural, namely it is simply
a subset of the given set of the poset such that any relation that
holds in the subposet also must hold in the given poset. We will discuss
this better next week.
- Homework for Tuesday, February 27. :
If P is a subposet of Q and they are both
posets on the same underlying set X, then we say that Q
is an extension of P. A partial order is a linear
order or a total order if any two elements are comparable.
This means that the Hasse diagram looks like a line, where one element
comes after another in a linear fashion. A linearly ordered set L
is a linear extension of a poset P if (i) L
is a linearly ordered set and
(ii) P is a subposet of L.
An interesting fact is that if a poset P contains two incomparable
elements x and y, then there is a linear extension
of P in which x is less than y. This
implies that any poset is, in fact, the intersetion of all of its linear
extensions, simply since if two elements x and y
are incomparable in the poset, then there are two linear extensions,
one in which x is less than y and another one
in which y is less than x. In particular, there
is a smallest number of linear extensions of a given poset, the intersection
of which will yield back the poset. This smallest number of linear extensions
is call the dimension of the poset. In general it is difficult
to compute the dimension of a given poset. A subset of the underlying set of
a poset that is linearly ordered is called a chain of the poset
and a subset of the underlying
set of a poset, every two of which are incomparable,
is called an antichain. If r is the number of elements
in the longest chain in a poset, then the underlying set of the posed can be
partitioned into exactly r antichains. Dually, if m
is the number of elements in the longest antichain in the poset, then
the underlying set can be partitioned into m chains.
This is Dilworth's Theorem and is harder to prove than
the partition into antichains.
- Homework for Tuesday, March 6. :
A particularly nice corollary of the partition of a poset into
antichains and chains is that: (i) if a poset can be partitioned
into r antichains, then r is at least the
size of the longest chain in the poset, (ii) if a poset can be
partitioned into m chains, then m is at least
the size of the largest antichain in the poset. There are both
direct consequences of the PHP. -- In particular, the poset
Pn of all the subsets of the set of the first
n natural numbers ordered by inclusion, can be partitioned
into n+1 antichains, and also by Sperner's Theorem
into m chains, where m is the size of the largest
antichain in Pn. This can have applications for
some questions in elementary number theory.
- Homework for Tuesday, March 20. :
By allowing arbitraty real powers when considering powers of x+1
we obtain Newton's Binomial Theorem (NBT) which allows us to obtain
many more identities involving binomial coefficients. Also, by considering
the sum of r variables instead of just two, we obtain the
Multinomial Theorem (MT), which also is fairly easy to prove.
Further, many familiar formulae that holds for the binomial coefficients
hava an analogus form for the more general multinomial coefficients, for example
the Pyramidal Pascal's identity for three variables, that resembles
the two dimensional Pascal's triangle in the plane.
Here is a nice wikipedia blurb on the
and the very layers I was cribbling on the board in considerable haste during
the last lecture. As you can see, the three-sided pyramid has three sides
consisting of Pascal's Triangle.
- Homework for Tuesday, March 27. :
The Inclusion/Exclusion Principle (I/E) is an important method
on how to count the number of elements of a union of sets, where they
can intersect in an arbitrary fashion. For two and three sets the formula
is fairly simple, but for k sets, the number of summands is
2k. In many cases it is convenient to count
the number of elements that don't have any of the given k
properties. Using I/E we were able to demonstrate how to compute the
number of r-combinations of a general finite multiset.
This we did by translating the problem to that of counting the number
of integer solutions of an additive equation where the variables have
each certain restrictions.
- Homework for Tuesday, April 3. :
We used the I/E to find a formula for derangements of
the integers 1,2,...,n. We found both an explicit formula
and also two recurrence relation. One interesting fact about the
number of derangements is that the ratio of the number of
derrangements to that of all permutations tends to 1/e
where e is the base number of the natrual logarithm.
We also saw that the number of derangements is a special case
of a more general number that tallies the number of permuations
where for each i it cannot be contained in given
set Xi. Determining such permutations
can be difficult, but by translating this problem into the problem
of placing rooks onto a board, we can in many special cases compute them
exactly, especially if the forbidden positions are not too "bad".
- Homework for Tuesday, April 10. :
In addition to compute the number of permutations where certain
absolute positions are forbidden, we can also use the I/E to compute
the number of permutations with certain relative forbidden
positions. We derived a formula for one such type of permutations
where the pattern i,i+1 did not occur for any
i. -- When considering recurrences and generating
functions, a good place to start is with the Fibonacci Numbers
0,1,1,2,3,5,8,13,21,34,..., where each term is the sum
of the previous two. As we briefly discussed, it depends on the task
at hand what is better; the recursive definition or the formula
for them that involves the
If the task is to print out the first million terms, then we should
definitely use the defining recursion, but if we need to examine the behavior
of the n-th term when n tends to infinity, then
it is definitely better to use the closed formula. We also presented
an alternate formula for the n-th Fibonacci Number, involving
binomial coefficients, all of which are integers, but the downside is
that we need at least (n+1)/2 summands. We will prove this formula
- Homework for Tuesday, April 17. :
Generating functions are used in a variety of ways in combinatorics.
One of the main application of generating functions is that they
can provide us with methods to obtain closed formulas for discrete
number sequences that either are defined purely combinatorially or
recursively. This opens up a gateway of useful tools we already
know from calculus and function theory that we now we can use
in combinatorial settings. Of particular value is when a generating
function defines a "smooth" function on an open real/complex open
set, since then we can both differentiate the generating function
and/or integrate it to obtain additional information in our investigations.
As an example we saw how NBT provided a generating function for the
generalized binomial coefficients. In another example we saw how
we could with relative ease (i.e. purely algebraic manipulation) obtain
a formula for the number of integer solutions to an integer equation
involving four integer variables where each variable had a predescribed
range of integer values. Another application of generating functions
occurs when we are unable to find a formula for terms of our discrete
number sequence, but we are able to obtain a closed formula for
the corresponding generating function. In this way, we could obtain
each term explicitly from the function by taking the appropriate
derivatives. This is the next best thing to obtaining a formula
for the terms of the sequence. -- Next week we discuss in more detail
how to obtain the generating function for a recursively defined
number sequence, like the Fibonacci numbers and Lucas numbers etc.
- Homework for Tuesday, April 24. :
We saw how we could use generating functions to obtain an explicit
formula for the Fibonacci numbers in terms of powers of some irrational
numbers. We then generalized this method to show that any number sequence
that satisfies a linear recurrence of length k can be
solved (theoretically!) in the same fashion, by (1) obtaining the generating
function as a rational function where the denumerator is a polynomial
of degree k, (2) factoring this polynomial in the denumerator,
and then (3) use partial fractions techniques and the NBT
to obtain a form/formula
for the number sequence. This can, for sure, be a daunting task, but
we saw how this can be done in a mechanical way. -- We then started
discussing a very interesting number sequence in combinatorics,
the Catalan Numbers. These numbers are defined by a
very special recursion and they pop up in a variety of combinatorial
questions. Using generating functions we were able to obtaina very
nice compact formula for the Catalan Numbers. We will continue
to discuss some of their properties next week.
Here below is a review sheet for the midterm to be
held on Thursday, April 5. -- Please
note that this sheet contains questions that I could potentially
ask on the midterm, but it is not a recipe for the midterm!
The midterm will consist of 5 - 8 problems each involving some
computation or a short proofs of a statements. Basically,, the midterm
will be on stuff that we have covered up to and including March 27,
Chapter 2 sections 1 -- 5,
Chapter 3 sections 1 -- 3,
Chapter 4 section 5,
Chapter 5 sections 1 -- 6,
Chapter 6 sections 1 -- 3.
Below are some drafts of solutions for some of the problems
on the review sheet for the midterm. Please note that these
are meant to present you the main ideas, but are not all
fully written detailed solutions! -- Do attempt the problems
here above first, before you read and study the solutions.
Below are handwritten solutions to the midterm exam, as I
wanted to see them. The recorded (elevated) median score and
the recorded average score were both
60/100. Note that I record only the elevated/yellow number grade
and not the curve I gave in class. -- Please, read over
these solutions and compare to what you did yourselves.