<s>[INST] Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$ Q: Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$.
My working:
$(x+x^2+x^3+...)^3$
$=x^3(1+x+x^2+...)^3$
$= x^3 \sum_{r=0}^\infty \begin{pmatrix} {r+3-1}\\{r}\end{pmatrix}x^r$
$= \sum_{r=0}^\infty \begin{pmatrix} {r+2}\\{r}\end{pmatrix}x^{3+r}$
Hence the number of ways is $\begin{pmatrix} {2n-1}\\{2n-3}\end{pmatrix}$=$\begin{pmatrix} {2n-1}\\{2}\end{pmatrix}$. 
However, the actual answer is $\begin{pmatrix} {2n+2}\\{2}\end{pmatrix}$$-3$$\begin{pmatrix} {n+1}\\{2}\end{pmatrix}$. Did i go wrong somewhere in my proof? Or is it some conceptual understanding gone wrong? Thanks!
 [/INST] Since we are selecting $2n$ balls from a total of $3n$ balls, 
we can count how many ways we can choose the $n$ balls we are leaving out.
This is the number of solutions in nonnegative integers to $a+b+c=n$,
which is given by $\color{blue}{\dbinom{n+2}{2}}$.

Alternate solution:
We want to find the number of solutions to $x_1+x_2+x_3=2n$ where $0\le x_i\le n$ for each $i$.
If S is the set of all solutions in nonnegative integers and if $A_i$ is the set of solutions with $x_i>n$,
we have $\big|\overline{A_1}\cap\overline{A_2}\cap\overline{A_3}\big|=\big|S\big|-\big|A_1\big|-\big|A_2\big|-\big|A_3\big|=\color{blue}{\dbinom{2n+2}{2}-3\dbinom{n+1}{2}}$
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