How many binary words of length n are there in which 0 and 1 occur the same number of times and in which no two 0's are adjacent? I understand that, in order to satisfy the first two conditions (length n, same number of 0's and 1's) all that needs to be done is
$$ \frac {n!} { \frac {n} {2}! \times \frac {n} {2}!} $$
but I'm not sure how to account for the third condition (no consecutive 0's).

Not many. If you have too many consecutive 1s, there will be more 1s than 0s in the word. So we can count them:


*

*If the word starts with a 1, the only option is 1010...1010.

*If the word starts with a 0 and ends with a 1, the only option is 0101...0101.

*If the word starts and ends with a 0, we need to have one pair of 1s inside. These can start on all even positions except the last one (e.g. 011010, 010110), which is on $n/2-1$ positions.
So the final solution is $n/2+1$ (for even $n$; for odd $n$ the answer is of course 0).
