1 1.6 $1447,1005,1231$ These three numbers have many similarities: 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?
[Solution] First, consider the positive integers with the same digit being 1. Since the first digit is 1, the other 1 has 3 positions to choose from. The other two digits cannot be 1 and must be different, so there are $P_{9}^{2}$ different arrangements. Therefore, we have
$$
m_{1}=3 P_{9}^{2}=216 .
$$

Next, consider the positive integers with the same digit not being 1. In this case, there are 9 different ways to choose the same digit, and these two identical digits have 3 different arrangements in the last 3 positions. The other digit cannot be 1 and cannot be the same as the identical digit, so there are 8 different choices. Therefore, we have
$$
m_{2}=9 \times 3 \times 8=216 \text {. }
$$

In summary, the total number of four-digit numbers that meet the requirements is $m=m_{1}+m_{2}=432$.