jmc

The row 1, 1 has 2 numbers. The row 1, 2, 1 has 3 numbers. The row 1, 3, 3, 1 has 4 numbers. Each time we go down one row, we have one more number in the list. So, the row that starts 1, $k$ has $k+1$ numbers (namely, the numbers $\left(\genfrac{}{}{0ex}{}{k}{0}\right),\left(\genfrac{}{}{0ex}{}{k}{1}\right),\left(\genfrac{}{}{0ex}{}{k}{2}\right),\dots ,\left(\genfrac{}{}{0ex}{}{k}{k}\right)$.) So the row with 41 numbers starts $\left(\genfrac{}{}{0ex}{}{40}{0}\right),\left(\genfrac{}{}{0ex}{}{40}{1}\right),\left(\genfrac{}{}{0ex}{}{40}{2}\right),\dots$. The 39th number has two numbers after it, and it is the same as the number in the row with only two numbers before it (that is, the 39th number is the same as the 3rd). So, the 39th number is $\left(\genfrac{}{}{0ex}{}{40}{2}\right)=\frac{40\cdot 39}{2\cdot 1}=\boxed{780}$.