\section*{Problem 2 - 161242}

Given a natural number \(n \geq 1\).

Determine the number of different ways to distribute \(2 n\) red, \(2 n\) green, and \(2 n\) black balls into two containers \(Q_{1}\) and \(Q_{2}\) such that each container contains \(3 n\) balls.

Hint:

Two distribution possibilities are considered the same if and only if for each of the three colors, the number of balls of that color in \(Q_{1}\) is the same in both distribution possibilities (and consequently the same is true for \(Q_{2}\)).
}

If we assign to each distribution possibility the triple \(\left(x_{1}, x 2, x 3\right)\), where \(x_{1}, x_{2}\) and \(x_{3}\) denote the numbers of red, green, and black balls in \(Q_{1}\), then by assumption:

\[
x_{1}+x_{2}+x_{3}=3 n \quad ; \quad 0 \leq x_{1}, x_{2}, x_{3} \leq 2 n
\]

The number we are looking for is equal to the number of all triples of integers that satisfy (1). If \(x_{1}=k(k=\) \(0,1, \ldots, n)\), then from (1) we get:

\[
x_{2}+x_{3}=3 n-k \quad ; \quad 0 \leq x_{2}, \quad x_{3} \leq 2 n
\]

Since \(x_{3} \leq 2 n\), \(x_{2}\) can be exactly one of the numbers \(n-k, n-k+1, \ldots, 2 n\). In the case \(x_{1}=k(k=0,1, . . n)\), there are thus exactly \(n+k+1\) possibilities.

If \(x_{1}=n+k(k=1,2, \ldots, n)\), then from (1) we get:

\[
x_{2} \neq x_{3}=2 n-k \quad ; \quad 0 \leq x_{2}, x_{3} \leq 2 n
\]

Since \(x_{3} \geq 0\), \(x_{2}\) can be exactly one of the numbers \(0,1, \ldots, 2 n-k\). In the case \(x_{1}=n+k(k=1,2, \ldots, n)\), there are thus exactly \(2 n-k+1\) possibilities.

Since \(0 \leq x_{1} \leq 2 n\), there are no further cases for \(x_{1}\). For the number \(A\) of all possibilities, we thus get:

\[
\begin{aligned}
A & =\sum_{k=0}^{n}(n+k+1)+\sum_{k=1}^{n}(2 n-k+1) \\
& =\frac{(2 n+1)(2 n+2)}{2}-\frac{n(n+1)}{2}+\frac{2 n(2 n+1)}{2}-\frac{n(n+1)}{2} \\
& =(2 n+1)(n+1)-n(n+1)+n(2 n+1)=3 n^{2}+3 n+1
\end{aligned}
\]

The number of different possibilities is \(3 n^{2}+3 n+1\).

\section*{Adapted from [5]}