Your solution is incorrect because you’re counting some arrangements more than once. Suppose that the balls are numbered $1$ through $n$. Consider the arrangement that has balls $1,2,3$, and $4$ in the first bucket and every other ball in the second bucket. You’ll count this once when you put balls $1,2$, and $3$ in the first bucket initially. You’ll count it again when you put balls $1,2$, and $4$ in the first bucket initially. And you’ll count it two more times, once when the ‘insurance’ balls are $1,3,4$ and once when they’re $2,3,4$. You end up counting that one arrangement four times. An arrangement with $5$ balls in the first bucket gets counted $\binom53 = 10$ times, and it gets worse from there. Try robjohn’s hint: count the bad distributions instead of the good ones.