$1 \cdot 2$ If an ordered pair of non-negative integers $(m, n)$ does not require a carry when adding $m+n$ (in decimal), then it is called "simple". Find the number of all simple ordered pairs of non-negative integers whose sum is 1492.
［Solution］Since there is no carry in the sum at each digit, there are 3 ways to add up to 2 at the unit place:
$$
0+2,1+1,2+0 .
$$

There are 10 ways to add up to 9 at the tens place:
$$
\begin{array}{l}
0+9,1+8,2+7,3+6,4+5, \\
5+4,6+3,7+2,8+1,9+0 .
\end{array}
$$

Similarly, there are 5 ways to add up to 4 at the hundreds place and 2 ways to add up to 1 at the thousands place.
In summary, by the multiplication principle, there are $3 \times 10 \times 5 \times 2=300$ pairs of numbers that meet the requirement.