Math 4707 (Spring 2011)
Exam 3
Problem 1. [15 points] Is there a bipartite graph with degrees 4, 4, 4, 4, 4, 10, 12, 12, 12, 20, 20, 20?
Solution. No. This is similar to a homework problem. (Alternati
MATH 4707 MIDTERM I
March 3, 2010
INSTRUCTOR: Anar Akhmedov
Name:
Signature:
ID #:
Show all of your work. No credit will be given for an answer without some work or explanation.
Problem
Points
1
2
3
4
MATH 4707, HOMEWORK II, SOLUTIONS TO GRADED PROBLEMS
February 25, 2010
INSTRUCTOR: Anar Akhmedov
(1) 4.3.6
Solution: Let an denote the number of all such subsets of cfw_1, 2, , n. We will nd a
recurre
MATH 4707, HOMEWORK I, SOLUTIONS TO GRADED PROBLEMS
February 19, 2010
INSTRUCTOR: Anar Akhmedov
(1) 1.8.21
Solution: Let us count all possible subsets Y of A such that |B Y | = 1 There are
k dierent w
Combinatorial Arguments
A combinatorial argument, or combinatorial proof, is an argument that involves counting. We have already seen this type of argument, for example in the section on Stirling
numb
Math 4707
Inclusion-Exclusion and Derangements
Dennis White
October 18, 2010
Suppose we have a set S of objects and a list of subsets of S , A1 , A2 , . . . , Am . We would like to
count the number of
Math 4707: Introduction to
Combinatorics and Graph Theory
Lecture Addendum, February 18, 2013
Recurrence Relations and a Glimpse of Generating Functions
Given an innite sequence of numbers, a generati
Math 4707: Introduction to
Combinatorics and Graph Theory
Lecture Addendum, March 13th and 25th, 2013
Counting Closed Walks and Spanning Trees in Graphs
via Linear Algebra and Matrices
1
Adjacency Mat
Math 4707
The Catalan Nunbers
Dennis White
November 3, 2010
1
Introduction
The Catalan numbers are a remarkable sequence of numbers that solve a number of seemingly
unrelated counting problems. Richar
Math 4707
The Matching Theorem
Dennis White
November 3, 2010
A matching in a bipartite graph is a subset of the edges with no common vertices. Alternatively,
it is a 1-regular subgraph. Here is an exa
Math 4707
Chromatic Polynomials
Dennis White
November 3, 2010
The chromatic polynomial of a simple graph G, CG (), is the number of ways of properly coloring
the vertices of G using colors. For exampl
Math 4707: Introduction to
Combinatorics and Graph Theory
Lecture Addendum, March 13th and 25th, 2013
Counting Closed Walks and Spanning Trees in Graphs
via Linear Algebra and Matrices
1
Adjacency Mat
Thus, A(G)k = (A(G)k1 A(G)ij = bi1 a1j + bi2 a2j + + bin anj . Since a walk of length k must be at a vertex
ij
r after k 1 steps, we note that
bir arj = # of walks of length exactly k between vi and v
where P is some invertible n-by-n matrix. Call this diagonal matrix in the middle D.
As a consequence, the kth power of M can be written as
M k = P DP 1 P DP 1 . . . P DP 1 = P D k P 1 .
The kth power
2
Counting spanning trees in graphs
We next use linear algebra for a dierent counting problem. Recall that a spanning tree of a graph G is a
subgraph T that is a tree that uses every vertex of G. Befo
thus it suces to compute W11 (k) and W12 (k), noting that the quantity 3W11 (k) equals the total number of
closed walks in C3 . Considering the last possible edge of a closed walk from v1 to v1 , and
MATH 4707, HOMEWORK III, SOLUTIONS TO GRADED PROBLEMS
April 10, 2010
INSTRUCTOR: Anar Akhmedov
(1) 6.10.6
Solution: Since k divides n, we have n = ks for some integer s. an 1 = aks 1 =
ks
(a ) 1 = (ak
MATH 4707 MIDTERM II
April 14, 2010
INSTRUCTOR: Anar Akhmedov
Name:
Signature:
ID #:
Show all of your work. No credit will be given for an answer without some work or explanation.
Problem
Points
1
2
3
Math 4707 Lecture Notes 1.8.34 We need to divide this problem into cases because given the placement of the first two people chosen, the number of ways to choose a third person who is not their neighb
Solutions to selected problems in Homework 3
Exercise 5.4.5 We ip a coin n times (n 1). For which values of n are the
following paris of events independent?
(b) The rst coin ip was head; the number of
Math 4707: Introduction to Combinatorics and Graph Theory
Spring 2012
Exam II
Instructions
Check that this booklet consists of 8 pages.
To get full credit, you must show all steps in your solutions
Math 4707 (Spring 2012)
Exam 1
Problem 1. [10 points] There are n people sitting around a circular table. How many ways are there to choose
3 people, no two of whom are neighbors? We assume that n 6.
Math 4707: Introduction to Combinatorics and Graph Theory
Spring 2012
Exam I
Instructions
Check that this booklet consists of 8 pages.
To get full credit, you must show all steps in your solutions.
Solutions to selected problems in Homework 2
Exercise 4.3.9 (d) Prove the following identity:
n
n
n
F1 +
F2 + +
Fn+1 = F2n+1
0
1
n
Proof. We will give a combinatorial proof.
Recall that Jn = Fn+1 is t
Math 4707 (Spring 2011)
Exam 3
Problem 1. [15 points] Is there a bipartite graph with degrees 4, 4, 4, 4, 4, 10, 12, 12, 12, 20, 20, 20?
Solution. No. This is similar to a homework problem. (Alternati
Math 4707: Introduction to Combinatorics and Graph Theory
Spring 2011
Exam III
Instructions
Check that this booklet consists of 8 pages.
To get full credit, you must show all steps in your solutions
Math 4707: Introduction to Combinatorics and Graph Theory
Spring 2011
Exam II
Instructions
Check that this booklet consists of 8 pages.
To get full credit, you must show all steps in your solutions
Problem 1. [10 points] We have three sets A, B , and C such that |A| = n, |B | = p, |C | = q , B C A, and
B C = . Find the number of subsets X A such that |B X | = 1 and |C X | = 2.
Solution. X has 1