Given a multiset $S$ of $O$ ones and $Z$ zeros, I'd like to count the number of permutations of $S$ that when partitioned into length $T$ segments have at least one segment that is all ones. ($T$ must of course divide $O+Z$).
For example, given the multiset $\{1,1,1,1,1,0,0,0,0,0,0,0,0,0,0\}$ with 5 ones and 10 zeros, the number of permutations when partitioned into segments of 3 (e.g., one would be $\{\{1,1,1\},\{1,1,0\},\{0,0,0\},\{0,0,0\},\{0,0,0\}\}$) that have at least one $\{1,1,1\}$ is 330.
After messing around with the problem for a while I came to what appeared to work:
$\frac{(t-1)! (o+z) \binom{o}{t} (o-t+z)!}{o! z!}$ for $o,z,t$ ones, zeros, and partition length.
However, I then found that only works for cases where the number of ones is < twice the partition size.
I'm thinking I need to use inclusion-exclusion to extend this to work for all cases, but I'm not sure how to proceed.
Denote by $N=O+Z$ the number of elements and by $M=N/T$ the number of segments. If you fill $k$ particular segments, you have $\binom{N-kT}Z$ choices left for the zeros. Thus by inclusion-exclusion there are
$$ \sum_{k=1}^M(-1)^{k+1}\binom Mk\binom{N-kT}Z $$
admissible arrangements.