Example 2 (2000 National High School Competition Question) If: (1) $a, b, c, d$ all belong to $\{1,2,3,4\}$; (2) $a \neq b$, $b \neq c, c \neq d, d \neq a$; (3) $a$ is the smallest value among $a, b, c, d$. Then the number of different four-digit numbers $\overline{a b c d}$ that can be formed is $\qquad$
Solve by filling in 28. Reason: (1) When $\overline{a b c d}$ contains 4 different digits, it is clear that $a=1$. There are $3!=6$ such four-digit numbers.
(2) When abcd contains only 3 different digits, choosing 3 different digits from $1,2,3,4$ has $C_{4}^{3}$ methods. At this time, the value of $a$ is uniquely determined ($a$ is the smallest of the 3 chosen numbers). When $a=c$, $b, d$ have 2! ways to be arranged; when $a \neq c$, $b=d, c$ have 2 ways to be arranged, with $b=d$ being the remaining number. Therefore, the number of four-digit numbers at this time is $C_{4}^{3}(2+2)=16$.
(3) When $\overline{a b c d}$ contains only 2 digits, choosing 2 digits from $1,2,3,4$ has $C_{4}^{2}$ methods. At this time, $a=c$ is the smaller of the chosen numbers, with only one way to choose, and $b=d$ is the remaining number, hence the number of four-digit numbers at this time is $C_{4}^{2} \cdot 1 \cdot 1=6$.
In summary, by the principle of addition, the number of different four-digit numbers that meet the conditions is $6+16+6=28$.