Pumping lemma for $\{110^n10^{n-m}1^m: 1\le m<n\}$ How can I prove that the language $L=\{110^n 10^{n-m} 1^m \mid 1 \le m &lt; n \}$ is not regular using pumping lemma?

I chose the word $w = 110^{2p} 10^{2p-q} 1^q$ to prove the non-regularity. 
Then I started partitioning the word to $xyz$ with $|y| \ge 1$, $|xy| \le p$ and for each $i\ge 0$, $xy^iz$: 

\begin{align*}
x &amp;= 110^r\\
y &amp;= 0^s\\
z &amp;= 10^{2p - q} 1^q\,,
\end{align*}
where $r + s = p$
and $s &gt; 0$.

The second partition:
\begin{align*}
x &amp;= 1\\
y &amp;= 10^{2p}\\
z &amp;= 10^{2p-q}1^q\,.
\end{align*}

However, I got stuck at this point because I am not really sure how to find the other partitions and how the choose the correct $i$ for those which I already have. 
Any help with this will be appreciated.
You're basically there.  Taking $w=110^{2p}10^{2p−q}1^q$, there are actually three cases (you missed the first one).


$x=\varepsilon$, $y=110^{p-2}$, $z=0^{p+2}10^{2p-q}1^q$.
$x=1$, $y=10^{p-1}$, $z=0^{p+1}10^{2p-q}1^q$.
$x=110^r$ for some $r\in\{0, \dots, p-3\}$, $y=0^{p-r-2}$, $z=0^{p+r+2}10^{2p-q}1^q$.


In all three cases, taking $i=0$ means that the string $xy^iz$ is not in&nbsp;$L$.  I'd suggest trying $i=0$ and $i=2$ as your first two guesses in pumping lemma proofs&nbsp;&ndash; one of them usually works.

Notice that nothing above actually depends on the choice of&nbsp;$q$, so you can simplify things very slightly by just taking $q=1$ (or any other fixed, legal value).