18.314 | Fall 2014 | Undergraduate

# Combinatorial Analysis

Assignments

## Problem Set 7

Most of the problems are assigned from the required textbook Bona, Miklos. A Walk Through Combinatorics: An Introduction to Enumeration and Graph Theory. World Scientific Publishing Company, 2011. ISBN: 9789814335232. [Preview with Google Books]

A problem marked by * is difficult; it is not necessary to solve such a problem to do well in the course.

### Problem Set 7

• Due in Session 19
• Practice Problems
• Session 17: None from textbook
• Let n≥1, and let f(n) be the number of partitions of n such that for all k, the part k occurs at most k times. Let g(n) be the number of partitions of n such that no part has the form i(i+1), i.e., no parts equal to 2, 6, 12, 20, …. Show that f(n)=g(n). Use generating functions.
• Let f(n) denote the number of partitions of n with an even number of 1’s. Give a combinatorial proof and a generating function proof that f(n) + f(n-1) = p(n), the total number of partitions of n.
• Session 18: Chapter 8: Exercises 20, 21
• Problems Assigned in the Textbook
• Chapter 8: Exercises 27, 28, 32, 37*. In exercise 28, you can ignore the last sentence (about comparing with Exercise 4). Hint for 37. Consider the product 1/(1-qx)(1-qx²)(1-qx³)…
• (A8*) Show that the number of partitions of n for which no part appears exactly once is equal to the number of partitions of n for which every part is divisible by 2 or 3. For instance, when n=6 there are four partitions of the first type (111111,2211,222,33) and four of the second type (222, 33, 42, 6). Use generating functions.
• (A9) Show that the number of partitions of n for which no part appears more than twice is equal to the number of partitions of n for which no part is divisible by 3. For instance, when n=5 there are five partitions of the first type (5, 41, 32, 311, 221) and five of the second type (5, 41, 221, 2111, 11111). Use generating functions.
• Bonus Problems
• (B2) Find the generating function G(x) = Σ_n_≥0 anxn/n!,
where an+1 = (n+1)an-{n\choose 2}an-2 for n≥0, and a0=1.
Thus a1=1, a2=2, a3=5. You don’t need to find a formula for an.

## Course Info

Fall 2014
##### Learning Resource Types
Exams with Solutions
Problem Sets