Consider the polynomial
$$(x+2)^{2n} = (x^2 + 4x + 4)^n = \\ = (x^2 + x + x + x + x + 1 + 1 + 1 + 1)(x^2 + x + x + x + x + 1 + 1 + 1 + 1) \dots \\ (x^2 + x + x + x + x + 1 + 1 + 1 + 1)$$
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,
$$\binom{2n}{n} \cdot 2^n.$$