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*}
&amp;\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\
&amp;\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\
&amp;\qquad=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}\\
&amp;\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&amp;=1+1+6\\
&amp;=1+2+5\\
&amp;=1+3+4\\
&amp;=2+2+4\\
&amp;=2+3+3
\end{align*}
Other way at looking the problem.p(n,3) can be very useful.Idea behind is to put the answer as an arithmatic progression.p(n,3)=sum of all positive terms of (n-3-6t) and adding 1 if any term is zero,where t=0,1,2,3,4,5•••
Ex
p(11,3)=(11-3-6*0)+(11-3-6*1)=8+2=10;
p(12,3)=(12-3-6*0)+(12-3-6*1)=9+3=12;
p(15,3)=(15-3-6*0)+(15-3-6*1)+(15-3-6*2);
             In the last example we encounter the third term being zero.We have to add 1 for it.
Hence,
p(15,3)=12+6+0+1=19