8. There are 10 cards, each card has two different numbers from $1,2,3,4,5$, and no two cards have the same pair of numbers. These 10 cards are to be placed into five boxes labeled $1,2,3,4,5$, with the rule that a card with numbers $i, j$ can only be placed in box $i$ or box $j$. A placement is called "good" if the number of cards in box 1 is more than the number of cards in any other box. The number of "good" placements is $\qquad$.
There are 10 types of cards: $12,13,14,15,23,24,25,34,35,45$. Since on average 2 cards are placed in each box, Box 1 can only contain 3 or 4 cards.

If Box 1 contains 3 cards, let's assume they are $12,13,14$, then $15$ can only be placed in Box 5. At this point, the number of cards in the remaining boxes can only be $2,2,2,1$. If Box 5 only contains 1 card, then $25,35,45$ can only be placed in Boxes 2, 3, and 4. If Box 2 contains $23$, then $24,34$ can only be placed in Boxes 4 and 3; if Box 2 contains $24$, then $23,34$ can only be placed in Boxes 3 and 4. In this case, there are $C_{4}^{3} \cdot 2=8$ ways to place the cards.

If Box 5 contains 2 cards, let's assume they are $15,25$. At this point, if Box 2 contains 2 cards, $23,24$, then $35,45$ can only be placed in Boxes 3 and 4, and $34$ can be placed in either Box 3 or 4; if Box 2 only contains 1 card, let's assume it is $23$, then Box 3 contains $35,34$, and Box 4 contains $45,24$. In this case, there are $C_{4}^{3} \cdot C_{3}^{1}(2+2)=48$ ways to place the cards.

If Box 1 contains 4 cards, then the remaining 6 cards are $23,24,25,34,35,45$. In this case, each card has 2 ways to be placed and meets the requirements, so there are $2^{6}=64$ ways to place the cards.

In summary, there are $8+48+64=120$ "good" ways to place the cards.

Note: When Box 1 contains 4 cards, if the number of cards in each box is divided as $3,2,1,0$, there are $C_{4}^{1} \cdot C_{3}^{1} \cdot 2=24$ ways; if divided as $3,1,1,1$, there are $C_{4}^{1} \cdot 2=8$ ways; if divided as $2,2,1,1$, there are $C_{4}^{2}(1+1+2)=24$ ways; if divided as $2,2,0,0$, there are $C_{4}^{1} \cdot 2=8$ ways, totaling 64 ways.