Czech-Polish-Slovak Match

Consider the polynomial $$\begin{gathered} (x+2{)}^{2n}=(x^2+4x+4{)}^n= \\ =(x^2+x+x+x+x+1+1+1+1)(x^2+x+x+x+x+1+1+1+1)\dots \\ (x^2+x+x+x+x+1+1+1+1) \end{gathered}$$ and suppose that we multiply out the brackets, obtaining $9^n$ summands. We show the one-to-one correspondence between the number of $x^n$'s and the number of the distributions we deal with in the problem. Suppose we have such a distribution. If the $k$-th ball goes to $A$, we pick $x^2$ from the $k$-th bracket. If it goes to $B$, $C$, $D$, $E$, we pick the first, second, third or fourth $1$, respectively, from the $k$-th bracket. If it goes to $F$, $G$, $H$, $I$, then we take the first, second, third or fourth $x$ from the $k$-th bracket. Now if we multiply the factors we have chosen, we see, that the result is equal to $x^n$ if and only if $A$ gets the same number of balls as $B$, $C$, $D$, $E$ jointly. Therefore, the number of the distributions we are interested in is equal to the coefficient at $x^n$ in the polynomial $(x+2{)}^{2n}$, that is, $$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\cdot 2^n.$$