5-character passwords

The number of ways in which first two characters each can be chosen from 26 letters is $\displaystyle 26\choose 1$ .

The last three characters can be chosen from the digits $\displaystyle \{0,1, \ldots, 9\}$ in $\displaystyle 10\choose 3$ ways.

And their various permutations will amount to

$\displaystyle 3! \times {10 \choose 3}$ .

Therefore, total number of possible passwords is

$\displaystyle {26\choose 1} \times {26\choose1} \times 3! \times {10\choose 3}$

$\Rightarrow \boxed{486720}$