Partitions of $n$ with exactly 3 parts

Question

I'm learning about generating functions, and came across this question: find the generating function for the number of partitions of a number $n$, into exactly 3 parts. I just solved a problem where the condition was that each part is no greater than 3, but I'm kind of stuck on this one. Any ideas? Thanks!

Answer 1

Hint: The following can be found somewhat more detailed in this answer.

A generating function for the number of partitions with exactly three parts is \begin{align*} &\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\ &\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\ &\qquad=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}\\ &\qquad=x^3+x^4+2x^5+3x^6+4x^7+\color{blue}{5}x^8+7x^9\cdots \end{align*}

Example: There are $\color{blue}{5}$ partitions of $8$ with three summands \begin{align*} 8&=1+1+6\\ &=1+2+5\\ &=1+3+4\\ &=2+2+4\\ &=2+3+3 \end{align*}

Answer 2

The two numbers are related to each other in a neat sort of way. Let's suppose, for example, that we want to partition $n = 10$ into three parts. One such partition is $10 = 5 + 3 + 2.$ We can represent this partition as follows:

X X X X X    5
X X X        3
X X          2

Now take this pattern of X's and "transpose" it, by reflecting it about a line starting at the upper left corner and running down & to the right (like you'd transpose a matrix):

X X X     3
X X X     3
X X       2
X         1
X         1

This is a different partition of $n = 10$: $10 = 3 + 3 + 2 + 1 + 1$. Note that because the original pattern had three rows, the new pattern has three columns; which means that no part of the new partition is greater than three. In fact, we can do this for any such partition of 10 into three parts; and given a partition of 10 such that the largest part is 3, we can find a corresponding partition of 10 into three parts. We conclude that the number of each type of partitions must be the same. This idea can then be generalized:

The number of partitions of $n$ into exactly $m$ parts is equal to the number of partitions of $n$ such that the largest part is exactly $m$.

So, if you can find the number of partitions of $n$ such that the largest part is exactly 3, then you can find the number of partitions of $n$ into exactly 3 parts.

FYI, the patterns that I used above are called Young diagrams (or sometimes Ferrers diagrams), and they pop up all the time in upper-level combinatorics.