jmc

The row that starts 1, 11 is the row $\left(\genfrac{}{}{0ex}{}{11}{0}\right),\left(\genfrac{}{}{0ex}{}{11}{1}\right),\left(\genfrac{}{}{0ex}{}{11}{2}\right),\cdots$. The $k^{\text{th}}$ number in this row is $\left(\genfrac{}{}{0ex}{}{11}{k-1}\right)$. (Make sure you see why it's $k-1$ rather than $k$ on the bottom.) Therefore, the $9^{\text{th}}$ number is $\left(\genfrac{}{}{0ex}{}{11}{8}\right)$. We have $$\left(\genfrac{}{}{0ex}{}{11}{8}\right)=\left(\genfrac{}{}{0ex}{}{11}{11-8}\right)=\left(\genfrac{}{}{0ex}{}{11}{3}\right)=\frac{11\cdot 10\cdot 9}{3\cdot 2\cdot 1}=\boxed{165}.$$