Ten, $1447, 1005, 1231$ have many things in common: they are all four-digit numbers, their highest digit is 1, and they each have exactly two identical digits. How many such numbers are there?
For the same digit being 1, since the highest digit is 1, the other 1 has three positions to choose from, and the remaining positions have $\mathrm{P}_{8}^{2}$ arrangements, i.e., $3 \cdot P_{0}^{*}=3 \times 72=216$ (ways).

For the same digit not being 1, there are 9 choices for the same digit, 8 choices for the other digit, and the other digit can be placed to the left, right, or in between the two identical digits, giving $3 \cdot 9 \cdot 8=216$ ways,
$\therefore$ there are a total of $216+216=432$ numbers that meet the requirements of the problem.