8.4. Try to answer: Among the following two types of ten-digit numbers, which type has a larger sum: the first type of numbers contains the adjacent digits 1 and 2; the second type of numbers contains the adjacent digits 2 and 1? Explain your reasoning.
8.4. Let the set of the first type of numbers be denoted as $A$, and the set of the second type of numbers as $B$. Since some ten-digit numbers contain both 12 and 21, they belong to both $A$ and $B$. After removing such ten-digit numbers, the remaining sets of the two types of numbers are denoted as $A^{\prime}$ and $B^{\prime}$, and their sums are denoted as $S_{1}$ and $S_{2}$, respectively. It is easy to see that each number $a$ in $A^{\prime}$ must contain the fragment $1 \cdots 12 \cdots 2$, i.e., a sequence of 1s (at least one) followed by a sequence of 2s. If we replace all 1s in this fragment with 2s and all 2s with 1s, we obtain a number $b$ that belongs to $B^{\prime}$, and $a < b$. Different numbers $a_{1} \neq a_{2}$ in $A^{\prime}$ will result in different numbers $b_{1}$ and $b_{2}$ in $B^{\prime}$. Therefore, the number of elements in $A^{\prime}$ is no more than the number of elements in $B^{\prime}$, and $S_{1} < S_{2}$.