Hello venkat sai
As i'm able to see , there is no need to add any relation b/w n and k. Question is complete.
Here we just have to apply our common sense. He is asking to arrange such that no two one are adjacent , and if there are many 1's compare to 0's then it can't possible to keep the 1's away from each other hence 0 ways. $\binom{n+1}{k}$ is a general answer which will take care about all cases.
When we say $\binom{n}{r}$ , it's common condition that $r\leq n$. You can't se;ect 4 things out of 3 things.