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 $\left(\genfrac{}{}{0ex}{}{24}{4}\right)$ and $\left(\genfrac{}{}{0ex}{}{24}{5}\right)$.

$$\left(\genfrac{}{}{0ex}{}{24}{4}\right)=\left(\genfrac{}{}{0ex}{}{23}{3}\right)+\left(\genfrac{}{}{0ex}{}{23}{4}\right)=1771+8855=10626$$ $$\left(\genfrac{}{}{0ex}{}{24}{5}\right)=\left(\genfrac{}{}{0ex}{}{23}{4}\right)+\left(\genfrac{}{}{0ex}{}{23}{5}\right)=8855+33649=42504$$

Now that we have $\left(\genfrac{}{}{0ex}{}{24}{4}\right)$ and $\left(\genfrac{}{}{0ex}{}{24}{5}\right)$, we can use Pascal's identity again to find $\left(\genfrac{}{}{0ex}{}{25}{5}\right)$.

$$\left(\genfrac{}{}{0ex}{}{25}{5}\right)=\left(\genfrac{}{}{0ex}{}{24}{4}\right)+\left(\genfrac{}{}{0ex}{}{24}{5}\right)=10626+42504=\boxed{53130}$$