Partitions of $n$ with exactly 3 parts 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! 

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*}
