jmc

We can use Pascal's identity $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ to find $\left(\genfrac{}{}{0ex}{}{15}{9}\right)$.

$$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{15}{8}\right)+\left(\genfrac{}{}{0ex}{}{15}{9}\right)& =\left(\genfrac{}{}{0ex}{}{16}{9}\right)\to \\ 6435+\left(\genfrac{}{}{0ex}{}{15}{9}\right)& =11440\to \\ \left(\genfrac{}{}{0ex}{}{15}{9}\right)& =5005\end{array}$$

We can use the identity again to find $\left(\genfrac{}{}{0ex}{}{15}{10}\right)$.

$$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{15}{9}\right)+\left(\genfrac{}{}{0ex}{}{15}{10}\right)& =\left(\genfrac{}{}{0ex}{}{16}{10}\right)\to \\ 5005+\left(\genfrac{}{}{0ex}{}{15}{10}\right)& =8008\to \\ \left(\genfrac{}{}{0ex}{}{15}{10}\right)& =3003\end{array}$$

Therefore, $\left(\genfrac{}{}{0ex}{}{15}{10}\right)=\boxed{3003}$.