I have a question regarding counting permutations of a multiset's elements. The problem is the following:

Given a multi-set $M=\{0^{m}, 1^{n-m}\}$ the number of all possible permutations of its elements is given by the formula $\frac{n!}{m!(n-m)!}$. How can I come up with a formula that gives the number of permutations in which every sequence of length $k+1$, where $k$ is even, contains at least $\frac{k}{2}$ $1$s.

E.g. for $n=6, m=3, k=2$ the permutation $101010$ should be counted, but the permutation $111000$ should not be counted.

I tried to employ the inclusion-exclusion principle, but was unable to come up with a solution to the problem. I would appreciate any ideas, references or solutions.

Thanks!

**UPDATE:**
Thank you @Max and @DavidCallan for your solutions and helpful discussions! I can now completely follow your answers. Indeed I have formulated the question vaguely. My goal is to find the following: I need a function $A(m,n,k)$, which basically gives the number of permutations of the multi-set $M=\{0^{m}, 1^{n-m}\}$ in which every sequence of length $k+1$ contains at least $\frac{k}{2}$ $1$s. $A$ will then be evaluated for even $k$s *only*. I need this because I would like to know what is the probability that a random permutation of the aforementioned multi-set satisfies the condition for different $k$s. Having this in mind I think @Max's solution can directly give me the answer I need.