The following question is from Arthur Engel's Problem-Solving Strategies: $2n$ objects each of three kinds are given to two persons, so that each person gets $3n$ objects. Prove that this can be done in $3n^2+3n+1$ ways.
I realize that this question might be a duplicate. I just wanted to upload my generating-functional approach as solution.
My Solution: So, there are 3 types of objects. Let $A(x)=1+x+x^2+ \cdots +x^{2n}$ be the generating function for choosing objects from any one type of object.
Since there are 3 types of objects, the ways to choose them is simply ${[A(x)]}^3$, which in turn is equal to:
\begin{align} \ (1+x+x^2+ \cdots +x^{2n})^3 & = (\frac{1-x^{2n+1}}{1-x})^3 \\ & = (1-x^{2n+1})^3(1-x)^{-3} \\ & = (1-x^{6n+3}-3x^{2n+1}+3x^{4n+2}) {{r+3-1} \choose {3-1}}x^r\\ & = (1-3x^{2n+1}) {{r+3-1} \choose {3-1}}x^r \\ & = (1-3x^{2n+1}) {{r+2} \choose {2}}x^r \\ & = {{r+2} \choose {2}}x^r - 3{{r+2} \choose {2}}x^{r+2n+1} \end{align}
Now, exponents of $x$ needs to equal to $3n$. Doing the needful gives: $${{3n+2} \choose {2}}x^r - 3{{n-1+2} \choose {2}}x^{n-1+2n+1}$$ $$={{3n+2} \choose {2}}x^r - 3{{n+1} \choose {2}}x^{3n}$$ Since, we want coefficients of $x^{3n}$, we get: $${{3n+2} \choose {2}} - 3{{n+1} \choose {2}}$$ Simplifying and calculating gives the desired answer:
$$3n^2+3n+1$$
Sirs, if there's some flaw in the answer or some 'technical' problem, please do point it out and help me rectify it.