here is an approach: We select $0\le 2k<n$ balls. Then choose $k$ balls among them for $A$. Rest of the $k$ balls may be distributed among $4$ people in $4^{k}$ ways. Remaining $n-2k$ balls may be distributed among rest $4$ people in $4^{n-2k}$ ways. Since, $k$ can range from $0$ to $\lfloor \frac{n}{2}\rfloor$, we get \[\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}\binom{n}{2k}\binom{2k}{k}4^{n-k}\]

Ok, good; but can you show this is equal to the much simpler expression $2^n\binom{2n}{n}$?

satori wrote: Rest of the $k$ balls may be distributed among $4$ people in $4^{k}$ ways. Remaining $n-2k$ balls may be distributed among rest $4$ people in $4^{n-2k}$ ways. What? If we have $B+C+D+E=k$, for nonnegative integers $B,C,D,E$, don't we have $\binom{k+3}{3}$ distributions?

The balls and boxes are both labeled. (The boxes, because they have explicit names; the balls, because the problem says so.)

Ah, missed "labeled balls". Thanks

Might still be interesting in the case of unlabeled balls (just not the same problem ). (Of course, the same approach works; you get instead the sum $\sum_k \binom{k}{3} \binom{n - 2k + 3}{k}$, which simplifies as one of two seventh-degree polynomials in $n$ depending on whether $n$ is even or odd.) The problem of simplifying satori's answer to the expression I gave above is still open in this thread.

Who can explain me, which is the correct solution for this nice Problem?