How many ways are there to distribute 10 balls into 5 distinct boxes that no two adjacent boxes are both empty? 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?
 A: 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.
A: 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.
$\binom{14}{4} - 4*\binom{9}{2} - 9*\binom{9}{1} - 5*\binom{9}{0}=771$
