jmc

From $E$ to $F$, it is 3 steps to the right and 1 step down, for a total of $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=\frac{4}{1}=4$ different paths. From $F$ to $G$, it is 2 steps to the right and 3 steps down, for a total of $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5\times 4}{2\times 1}=10$ different paths. So there are $4\times 10=\boxed{40}$ paths from $E$ to $G$ that pass through $F$.