What I got:
$$\binom{14}{4} - 4\times \binom{9}{2} - 3\times \binom{9}{1} - 2\times \binom{9}{0}=828$$
Basically, I found the total number of ways and subtract the cases where $2, 3,$ and $4$ adjacent boxes are empty.
I am given this question and I do not have the answer, can anyone confirm/correct me?
I'm not convinced by your handling of the forbidden allocations. Here is what I got:
$0$ empty boxes: Put a ball into each box and distribute the remaining $5$ balls arbitrarily. Makes ${9\choose4}=126$.
$1$ empty box: Choose this box in $5$ ways, put a ball into the remaining four boxes, and distribute the remaining $6$ balls arbitrarily over these four boxes. Makes $5\cdot{9\choose3}=420$.
$2$ empty boxes: Choose these boxes in ${5\choose2}-4=6$ ways, put a ball into the remaining three boxes, and distribute the remaining $7$ balls arbitrarily over these three boxes. Makes $6\cdot{9\choose2}=216$.
$3$ empty boxes: There is just one way to choose these three boxes. Put a ball into the boxes Nr. 2 and 4, and distribute the remaining $8$ balls arbitrarily over these two boxes. Makes ${9\choose1}=9$.
It follows that there are $771$ admissible allocations.
Your approach is almost ok except you oversee that if first two boxes are empty the last box may also be empty, but you calculate as it must have a ball in it.
For leaving 3 boxes empty you must pick 2 boxes and only picking oxoxo isn't valid, so $\binom{5}{2}-1$. For leaving 4 boxes empty you may pick any box.
$$\binom{14}{4} - 4 \cdot \binom{9}{2} - 9 \cdot \binom{9}{1} - 5 \cdot \binom{9}{0}=771$$
