498 passengers randomly board 6 carriages, the number of ways to board such that exactly two carriages are empty (no one boards) is

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498 passengers randomly board 6 carriages, 
the number of ways to board such that exactly two carriages are empty (no one boards) is
$4915\left(4^{8}-4 \cdot 3^{8}+6 \cdot 2^{8}-4\right)$ or 612360. First, find the number of ways to make carriages $5$ and $6$ empty, and carriages $1, 2, 3, 4$ all occupied. The number of ways for 8 people to board 4 carriages is $N=4^{8}$. Let $N_{i}$ denote the number of ways for the $i$-th carriage to be empty, $N_{i}=3^{8}(i=1,2,3,4)$. Let $N_{ij}$ denote the number of ways for the $i$-th and $j$-th carriages to be empty, $N_{ij}=2^{8}$. Let $N_{ijk}$ denote the number of ways for the $i$-th, $j$-th, and $k$-th carriages to be empty, $N_{ijk}=1$. The number of ways for all four carriages to be empty is 0, i.e., $N_{1234}=0$. By the principle of inclusion-exclusion, the number of ways for all four carriages to be occupied is $N_{0}=N-C \cdot 3^{8}+C_{4}^{2} \cdot 2^{8}-C_{4}^{8} \cdot 1=4^{8}-4 \cdot 3^{8}+6 \cdot 2^{8}-4$. Since there are $C_{6}^{2}=15$ ways to choose which two carriages are empty, the answer is $15\left(4^{8}-4 \cdot 3^{8}+6 \cdot 2^{8}-4\right)=612360$.