Combinatorics Exam 2, Exams of Algebra

An exam for the Combinatorics course at Carnegie Mellon University. It includes three problems with their respective solutions. The exam is closed book and students are not allowed to consult their notes, textbooks, other students or electronic equipment during the exam. The solutions require students to justify their answers and use known series expansions for functions without proof as long as they state what they are using. The exam covers topics such as subsets, edge colouring, and mono-coloured cliques.

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2014/2015

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Department of Mathematics
Carnegie Mellon University
21-301 Combinatorics
Section B
Exam 2 - 3rd March 2015
Name:
This is a closed book exam, you may not consult your notes,
textbooks, other students or electronic equipment during the
exam. You may use known series expansions for functions with-
out proof as long as you state what you are using. If you make
use of something we proved during lectures, be very explicit in
doing so by stating exactly what results or properties you are
using and why they apply. You may not cite without proof
theorems you proved on homework/review sheet or read in the
book/the internet/elsewhere. You must justify your answers.
Problem Points Score
1 30
2 35
3 35
Total: 100
1
pf3
pf4
pf5

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Department of Mathematics

Carnegie Mellon University

21-301 Combinatorics

Section B

Exam 2 - 3rd March 2015

Name:

This is a closed book exam, you may not consult your notes, textbooks, other students or electronic equipment during the exam. You may use known series expansions for functions with- out proof as long as you state what you are using. If you make use of something we proved during lectures, be very explicit in doing so by stating exactly what results or properties you are using and why they apply. You may not cite without proof theorems you proved on homework/review sheet or read in the book/the internet/elsewhere. You must justify your answers.

Problem Points Score 1 30 2 35 3 35 Total: 100

(a) Prove that (for 1 < k < n), ( n k

(n k

)k .

Solution ( n k

n(n โˆ’ 1)... (n โˆ’ k + 1) k(k โˆ’ 1)... 2 ร— 1

n k

n โˆ’ 1 k โˆ’ 1

n โˆ’ 2 k โˆ’ 2

n โˆ’ k + 1 k โˆ’ k + 1

For 1 โ‰ค i โ‰ค k โˆ’ 1 we have that nkโˆ’โˆ’ii > nk. This follows since the following are equivalent statements, n โˆ’ i k โˆ’ i

n k nk โˆ’ ik > nk โˆ’ in in > ik n > k,

and we know the final statement is true. This gives us that ( n k

(n k

) (n k

(n k

(n k

)k .

(b) Prove that 2 n n + 1

n bn/ 2 c

โ‰ค 2 n.

Solution

(n k

is the number of ways of choosing a subset of size k from a set of size n. We know that the number of ways of choosing a subset of any size from n elements is 2 n^ since each element of [n] can be either in or out of the subset. Therefore for all k,

(n k

< 2 n.

  1. You are organising a school sports day. The school will be split into two teams who will compete in many different sporting events. The students will decide what sports will be played and have formed 30 committees of 6 students each to discuss organisation and rules for each sport. To ensure the competition is fair, you must choose the two teams, and must make sure that each committee has at least one student from each team. Show that this is possible. Solution Call the teams A and B. Assign each student to A or B with probability half each, uniformly and independently at random. The probability that a committee is entirely on team A is therefore 216. Equally the probability that it is entirely on team B is the same. Therefore, since these events are distinct events on the same probability space, the probability that a committee is entirely on the same team is

2 ร—

= 2โˆ’^5 =

The expected number of committees with all members of one team is therefore 30ร— 321. Since this is clearly less than 1, and the number of committees with all members on one team can only take integer values, there must be a greater than 0 probability that this value is 0. This tells us that there must exist at least one assignment of students to teams such that the number of committees with all members on one team is 0 and this is the assignment we require.

  1. Let n^2 < 2 kโˆ’^1. Show that there exists an (edge) colouring of the complete graph on n vertices with two colours such that there exists no mono-coloured clique of size k (i.e. no k vertices where all the edges between those k vertices are one colour.) Solution Randomly colour the edges of the graph either red or blue independently of the colour of any other edge, with proba- bility half each. There are

(n k

cliques of size k and each clique contains

(k 2

edges. The probability that a clique is mono-coloured is 2ร—(1/2)(

k 2 ) , since there are two choices of colour for the mono-coloured clique to be and the probability that all of the edges are that colour is (1/2)(

k 2 ) . The expected number of mono-coloured cliques is therefore ( n k

k 2 )

n k

21 โˆ’^

k(k 2 โˆ’1)

n k

2 โˆ’^

k(kโˆ’1) 2

โ‰คnk 2 โˆ’^

k(k 2 โˆ’1)

n 2 โˆ’^

(kโˆ’ 2 1)^ )k

n^22 โˆ’(kโˆ’1)

)k 2 .

The third line follows from noting that,

n k

n(n โˆ’ 1)... (n โˆ’ k + 1) k(k โˆ’ 1)... 3 ร— 2 ร— 1

= n(n โˆ’ 1)... (n โˆ’ k + 1) k(k โˆ’ 1)... 3 ร— 1

โ‰ค nk.