Example 1. Using the ten digits from $0-9$,
(1) How many 3-digit numbers without repeated digits can be formed?
(2) How many of these numbers are between $300-700$ and do not end in zero?
(1) Since different arrangements of three digits form different numbers (ordered), this is a permutation problem. We can arrange them in the order of hundreds, tens, and units. The first digit cannot be 0, so there are $\mathrm{P}_{\mathrm{P}}^{1}$ ways to arrange the first digit. The next two digits can be arranged from the remaining 9 digits, taken two at a time, which gives $\mathbf{P}_{0}^{2}$ ways. We can draw a diagram: $\mathbf{P}_{\theta}^{1} \mathbf{P}_{9}^{2}$. Since these two steps must be completed to form a 3-digit number, they are related, and we use multiplication. This gives us $P$ . $P_{9}^{2}=648$ (ways).

(2) For numbers greater than 300 and less than 700, the first digit can only be 3, 4, 5, or 6, so there are $\mathrm{P}$ ! ways to arrange the first digit. For the last two digits, we need to consider that the last digit cannot be 0. If we exclude 0, there are $\mathbf{P}_{8}^{2}$ ways to arrange the last two digits, but this is incorrect because 0 can be in the middle but not at the end. Replacing it with $\mathbf{P}_{\mathbf{g}}^{2}$ is also incorrect, as this includes cases where 0 is at the end. For such problems, we should prioritize the special conditions. First, arrange the last digit, which has $\mathbf{P}_{8}^{1}$ ways to arrange after the first digit is arranged. After the first and last digits are arranged, the middle digit has $\mathrm{P}_{8}^{1}$ ways to arrange. The diagram is: \begin{tabular}{|l|l|l|} $\mathbf{P}_{4}^{1}$ & $\mathbf{P}_{8}^{1}$ & $\mathbf{P}_{8}^{1}$ \\ \hline \end{tabular}, and the final result is: $\mathbf{P}_{4}^{\frac{1}{4}} \cdot \mathbf{P}_{8}^{1} \cdot \mathbf{P}_{8}^{1}=256$ (ways).