Regular expression for binary words with few zeros
@SamM's answer isn't completely right. Yes, it is $(01^m)^*$ for "each zero is followed by exactly $m$ ones", but this can't just be combined with "at least $n$ ones before". And a string of only ones complies always. So consider the ones between two zeroes:


If $m &lt; n$, you can't comply with both conditions, so the language is just $1^n 1^* 0 1^m \mid 1^*$
If $m \ge n$, the only way to comply in between zeroes is to have $m$ ones there, so the language is $1^n 1^* (0 1^m)^+ \mid 1^*$