jmc

We can apply Pascal's identity to get that $\left(\genfrac{}{}{0ex}{}{19}{9}\right)=\left(\genfrac{}{}{0ex}{}{18}{8}\right)+\left(\genfrac{}{}{0ex}{}{18}{9}\right)$. From here, we can apply it twice more to get that $\left(\genfrac{}{}{0ex}{}{19}{9}\right)=\left(\genfrac{}{}{0ex}{}{18}{8}\right)+\left(\genfrac{}{}{0ex}{}{18}{9}\right)=\left(\left(\genfrac{}{}{0ex}{}{17}{7}\right)+\left(\genfrac{}{}{0ex}{}{17}{8}\right)\right)+\left(\left(\genfrac{}{}{0ex}{}{17}{8}\right)+\left(\genfrac{}{}{0ex}{}{17}{9}\right)\right)$. Substituting the provided values of $\left(\genfrac{}{}{0ex}{}{17}{7}\right)$, $\left(\genfrac{}{}{0ex}{}{17}{8}\right)$, and $\left(\genfrac{}{}{0ex}{}{17}{9}\right)$ gives us $\left(\genfrac{}{}{0ex}{}{19}{9}\right)=19448+2(24310)+24310=\boxed{92378}$.