jmc

There are 26 ways to choose the repeated letter, $\left(\genfrac{}{}{0ex}{}{25}{2}\right)$ ways to choose the other two letters, $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ways to choose in which two of the four positions to put the repeated letters, 2 ways to choose how to arrange the remaining two letters, 10 ways to choose the first digit and 9 ways to choose the second digit for a total of $(26)\left(\genfrac{}{}{0ex}{}{25}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)(2)(10)(9)=\boxed{8,424,000}$ combinations.