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I need to solve this problem:

How many ways are there,$2n$ ball $A$ and $2n$ ball $B$ and $2n$ ball $C$ distribute to two equal group so that every group has 3n member?

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In many combinatorial problems the question is, which of the dozen combinatorial formulas I have seen applies? This is not the case here. The task is to get to the heart of the situation, and then its simple counting.

I'm assuming that the two groups are labeled $1$ and $2$.

Group $1$ may take $a$ balls of type $A$ and $b$ balls of Type $B$ under the sole conditions that $$0\leq a\leq 2n,\quad 0\leq b\leq 2n,\quad n\leq a+b\leq 3n\ ;\qquad(*)$$ finally group $1$ should take as many balls of type $C$ as are needed to make a total of $3n$. When this has been done group $2$ should take all the remaining balls.

It follows that there are just two degrees of freedom: The choice of $a$ and $b$, whereby the restrictions $(*)$ apply. In order to count the number of possibilities satisfying $(*)$ we draw an $(a,b)$-plane and shade the $(a,b)$-domain $F$ where all six inequalities $(*)$ are fulfilled. Here is the figure:

enter image description here

Now we have count the number of integer points in $F$. The square $Q:=[0,2n]^2$ contains $(2n+1)^2$ lattice points, and each of the small triangles in the lower left and upper right corners of $Q$ contains $$1+2+\ldots+n={n(n+1)\over2}$$ lattice points. Therefore the number $f(n)$ of feasible lattice points $(a,b)$ is given by $$f(n)=(2n+1)^2 -2{n(n+1)\over2}=3n^2+3n+1\ .$$ In particular $f(1)=7$, as can be easily verified by going through the cases.

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I don't know how to count possibles?can you help more? – Sali Me Dec 14 '12 at 19:33
@Sali Me: See my edit, containing a figure. – Christian Blatter Dec 14 '12 at 19:48

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