1. Set $A=\{0,1,2,3,4,5,6,7\}$, how many mappings $f$ from $A$ to $A$ satisfy the following conditions (1) and (2)?
(1) $i, j \in A, i \neq j$, then $f(i) \neq f(j)$;
(2) $i, j \in A, i+j=7$, then $f(i)+f(j)=7$.
1. Solution: Let $A_{0}=\{0,7\}, A_{1}=\{1,6\}, A_{2}=\{2,5\}, A_{3}=\{3,4\}$. From condition (2), we know that the images of elements in $A_{i}$ must be in the same $A_{j}$. From (1), for different $A_{i}$, the corresponding $A_{j}$ are different. Thus, the ordered pairs of $i$ and $j(1 \leqslant i, j \leqslant 4)$ have 4! arrangements, and the elements in each $A_{j}$ as images can be swapped, hence there are $2^{4}$ ways. Therefore, the total number of mappings required is 4 $\times 2^{4}=384$ (ways).