Example 1 The numbers 1447, 1005, and 1231 have certain common points, that is, each number is a four-digit number starting with 1, and in each four-digit number, exactly two digits are the same. How many such four-digit numbers are there?
Solution: The qualified four-digit numbers must contain one 1 or two 1s.
(1) The case with two 1s: From the remaining 9 digits (excluding 1), any two can be chosen in $C_{9}^{2}$ ways, and then combined with one 1 to form any permutation of a three-digit number, which has $P_{3}^{3}$ ways. Thus, the number of four-digit numbers with the first digit being 1 is $N_{1}=C_{9}^{2} P_{3}^{3}$.
(2) The case with only one 1: From the remaining 9 digits, any two can be chosen in $C_{9}^{2}$ ways, and one of these digits is chosen to be repeated in $C_{2}^{1}$ ways. The three digits thus formed can be arranged in $P_{3}^{1}$ ways. Therefore, the number of four-digit numbers with the first digit being 1 is $N_{2}=C_{9}^{2} C_{2}^{1} P_{3}^{1}$, where $P_{3}^{1}$ indicates selecting one position for the non-repeating digit and the remaining two positions for the repeating digit. Hence, the total number of four-digit numbers that meet the criteria is
$$
N=N_{1}+N_{2}=432 \text { (numbers). }
$$