Example 1-15 Find the number of 4-digit numbers composed of different digits that are even, between 2000 and 7000.
Let this four-digit number be $n_{3} n_{2} n_{1} n_{0}$,
Since it is an even number, $n_{0}$ can only take the digits $0,2,4,6,8$.
Restricted to numbers between 2000 and 7000, $n_{3}$ can only take the digits $2,3,4,5,6$. We discuss the cases as follows:
(1) When $n_{3}$ takes 2, 4, or 6, $n_{0}$ can only have 4 choices, and $n_{2} n_{1}$ can only be chosen from the remaining 8 digits for 2 permutations, so the number of numbers that meet the condition, according to the multiplication rule, is
$$
3 \times 4 \times P(8,2)=12 \times 8 \times 7=672 \text { numbers }
$$
(2) If $n_{3}$ does not take 2, 4, or 6, then $n_{0}$ has 5 choices, and $n_{2} n_{1}$ is chosen from the remaining 8 digits for 2 permutations, so the number of even numbers that meet the condition is
$$
2 \times 5 \times P(8,2)=560 \text { numbers }
$$

According to the addition rule, the total number of even numbers is
$$
672+560=1232 \text { numbers }
$$