Regex for at least three 1's and at most four 0's
Here is a solution for at least $a$ 1’s and at most $b$ 0’s:
$$
\sum_{a_0+\cdots+a_b=a} 1^{a_0}1^*\prod_{i=1}^b (\epsilon+0)1^{a_i}1^*,
$$
where $a_0,\ldots,a_b \geq 0$ are integers. For example, when $a=b=2$ we get
$$
\begin{align*}
&amp;111^*(\epsilon+0)1^*(\epsilon+0)1^*+\\
&amp;1^*(\epsilon+0)111^*(\epsilon+0)1^*+\\
&amp;1^*(\epsilon+0)1^*(\epsilon+0)111^*+\\
&amp;11^*(\epsilon+0)11^*(\epsilon+0)1^*+\\
&amp;11^*(\epsilon+0)1^*(\epsilon+0)11^*+\\
&amp;1^*(\epsilon+0)11^*(\epsilon+0)11^*.
\end{align*}
$$
With more effort, you can even create an unambiguous regular expression, that is, one which corresponds to a UFA rather than an NFA; in other words, whenever there is a sum, the corresponding languages are disjoint. As an example, here is an unambiguous regular expression corresponding to $a=b=1$:
$$
1^*+11^*01^*+011^*.
$$
Compare this to the ambiguous one constructed above:
$$
11^*(\epsilon+0)1^*+1^*(\epsilon+0)11^*.
$$
The words $1,101$ are both captured in both summands above, but in only a single summand in the unambiguous regular expression.