Example 3 For the four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b, c, d \leqslant 9)$: If $a>b, bd$, then $\overline{a b c d}$ is called a $P$ class number; If $ac, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbers and $Q$ class numbers, respectively. Then the value of $N(P)-N(Q)$ is $\qquad$. [3] $(2015$, National High School Mathematics Joint Competition)
Let's denote the set of all numbers of type $P$ and $Q$ as $A$ and $B$ respectively, and the set of all numbers of type $P$ whose last digit is zero as $A_{0}$, and the set of all numbers of type $P$ whose last digit is not zero as $A_{1}$.

For any four-digit number $\overline{a b c d} \in A_{1}$, map it to the four-digit number $\overline{d c b a}$.
Notice that, $a > b$, $b < c$, $c > d \geqslant 1$.
Thus, $\overline{d c b a} \in B$.
Conversely, each $\overline{d c b a} \in B$ uniquely corresponds to an element $\overline{a b c d}$ in $A_{1}$.
Therefore, a one-to-one correspondence is established between $A_{1}$ and $B$.
Hence, $N(P) - N(Q) = |A| - |B|$
$= |A_{0}| + |A_{1}| - |B| = |A_{0}|$.
For any four-digit number $\overline{a b c 0} \in A_{0}$, $b$ can take values 0, 1, ..., 9. For each $b$, since $b < a \leqslant 9$ and $b < c \leqslant 9$, $a$ and $c$ each have $9 - b$ possible values. Therefore,
$$
|A_{0}| = \sum_{b=0}^{9} (9 - b)^2 = \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285.
$$

Thus, $N(P) - N(Q) = 285$.
[Note] In this problem, $N(P)$ and $N(Q)$ themselves can be calculated, but the above solution method, through appropriate pairing, allows for a significant reduction in the values of $N(P)$ and $N(Q)$ before calculating $N(P) - N(Q)$, thereby reducing the computational load.