Skip to main content
Logo image

In-Class Notes to Accompany Applied Combinatorics

Handout 8.2 Partitions and Newton’s Binomial Theorem

Partitions of Integers.

Definition 8.4.

A partition of the positive integer \(n\) is a way of writing \(n\) as a sum of nonincreasing positive integers.

Definition 8.5.

A partition of a positive integer is said to be a partition into odd parts if every term (also referred to as a part) in the sum is odd.

Definition 8.6.

A partition of a positive integer is said to be a partition into distinct parts if each integer appears in the sum at most once.

Activity 8.2.1.

List the partitions of \(9\text{.}\) Count the number of partitions into odd parts. Count the number of partitions into distinct parts.
Hint.
Be systematic! One way to do this is by grouping partitions by their largest part.
\begin{align*} 9\amp \amp 3\amp +3+3\\ 8\amp +1 \amp 3\amp +3+2+1\\ 7\amp +2 \amp 3\amp +3+1+1+1\\ 7\amp +1+1 \amp 3\amp +2+2+2\\ 6\amp +3 \amp 3\amp +2+2+1+1\\ 6\amp +2+1 \amp 3\amp +2+1+1+1+1\\ 6\amp +1+1+1 \amp 3\amp +1+1+1+1+1+1\\ 5\amp +4 \amp 2\amp +2+2+2+1\\ 5\amp +3+1 \amp 2\amp +2+2+1+1+1\\ 5\amp +2+2 \amp 2\amp +2+1+1+1+1+1\\ 5\amp +2+1+1 \amp 2\amp +1+1+1+1+1+1+1\\ 5\amp +1+1+1+1 \amp 1\amp +1+1+1+1+1+1+1+1\\ 4\amp +4+1 \amp \\ 4\amp +3+2 \amp \\ 4\amp +3+1+1 \amp \\ 4\amp +2+2+1 \amp \\ 4\amp +2+1+1+1 \amp \\ 4\amp +1+1+1+1+1 \amp \end{align*}
How would we write a generating function in which the coefficient on \(x^n\) is the number of partitions of \(n\text{?}\)

Activity 8.2.2.

(a)

Write a generating function in which the coefficient on \(x^{n}\) is the number of partitions of \(n\) into distinct parts.
Hint.
This is conveniently done as a product of simple generating functions.

(b)

Write a generating function in which the coefficient on \(x^{n}\) is the number of partitions of \(n\) into odd parts.
Hint.
A convenient form is a product of rational functions, but you might want to start with a product of power series and rewrite it.

(c)

Show that your generating functions above are actually equal to one another.
Hint.
\(\displaystyle (1+x^{k})\frac{1-x^{k}}{1-x^{k}}= \frac{\text{?}}{1-x^{k}}\)

Newton’s Binomial Theorem.

Definition 8.7.

For all real numbers \(p\) and nonnegative integers \(k\text{,}\) the number \(P(p,k)\) is defined by
  1. \(P(p,0)=1\) for all real numbers \(p\) and
  2. \(P(p,k) = p P(p-1,k-1)\) for all real numbers \(p\) and integers \(k\gt 0\text{.}\)

Definition 8.8.

For all real numbers \(p\) and nonnegative integers \(k\text{,}\)
\begin{equation*} \binom{p}{k}=\frac{P(p,k)}{k!}\text{.} \end{equation*}

Activity 8.2.3.

Compute \(\displaystyle\binom{-9/2}{5}\text{.}\)

Activity 8.2.4.

(a)
Use mathematical induction to show that for all \(k\geq 0\text{,}\) \(\displaystyle \binom{-1/2}{k}= (-1)^{k}\frac{\binom{2k}{k}}{2^{2k}}\text{.}\)
(b)
Use Newton’s Binomial Theorem and the step above to write \((1-4x)^{-1/2}\) as a formal power series in which the coefficient on \(x^{n}\) is a binomial coefficient in which both numbers are integers.