u上as $8$ 个乘客随意登上 6 节车相, 使恰有两节车厢空着的上车方法总数是多少?

There are $8$ passengers boarding 6 carriages randomly. How many ways are there for the passengers to board the carriages such that exactly two carriages are empty?
Assume that carriages $5$ and $6$ are empty, and carriages $1, 2, 3, 4$ are all occupied. Let the set of all ways for 8 people to board these 4 carriages be $S$, then $|S|=4^{8}$. Let $A_{i}$ represent the set of ways where the $i$-th carriage is empty $(i=1,2,3,4)$, then $\left|A_{i}\right|=3^{8}(i=1,2,3,4),\left|A_{i} \cap A_{j}\right|=2^{8}(1 \leqslant i< j \leqslant 4),\left|A_{i} \cap A_{j} \cap A_{k}\right|=1(1 \leqslant i<j<k \leqslant 1),\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\right|=0$. By the principle of inclusion-exclusion: $\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3} \cap \bar{A}_{4}\right|=|S|-\mathrm{C}_{4}^{1} \cdot\left|A_{1}\right|+\mathrm{C}_{4}^{2} \cdot\left|A_{1} \cap A_{2}\right|-\mathrm{C}_{4}^{3} \cdot \left|A_{1} \cap A_{2} \cap A_{3}\right|+\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\right|=4^{8}-4 \cdot 3^{8}+6 \cdot 2^{8}-4=40824$, so the total number of boarding ways is $C_{6}^{2} \cdot 40824=612360$.