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. Place these 10 cards 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 and only if the number of cards in box 1 is more than the number of cards in each of the other boxes. How many good placements are there?
8. 120.

Use $\{i, j\}$ to denote the card with $i$ and $j$ written on it.
It is easy to see that these 10 cards are exactly $\{i, j\}(1 \leqslant i < j \leqslant 5)$.
Consider the good ways to place the cards.
There are 10 cards in total in the five boxes, so box 1 must have at least three cards. The cards that can be placed in box 1 are only $\{1,2\}, \{1,3\}, \{1,4\}, \{1,5\}$.
(1) All four of these cards are in box 1. In this case, it is impossible for any of the remaining boxes to have 4 cards, so the remaining 6 cards can be placed in any way, resulting in $2^6 = 64$ good placements.
(2) Exactly three of these four cards are in box 1, and the remaining cards in each box are at most two.

Consider the number of placements $N$ where $\{1,2\}, \{1,3\}, \{1,4\}$ are in box 1, and $\{1,5\}$ is in box 5.

The cards $\{2,3\}, \{2,4\}, \{3,4\}$ have eight possible placements, six of which involve placing two cards in one of the boxes 2, 3, or 4, and the remaining two involve placing one card in each of the boxes 2, 3, and 4.

If two of $\{2,3\}, \{2,4\}, \{3,4\}$ are in the same box, say $\{2,3\}, \{2,4\}$ are in box 2, then $\{2,5\}$ must be in box 5, which already has $\{1,5\}$ and $\{2,5\}$. Therefore, $\{3,5\}$ and $\{4,5\}$ must be in boxes 3 and 4, respectively, meaning the placement of $\{2,5\}, \{3,5\}, \{4,5\}$ is unique;

If $\{2,3\}, \{2,4\}, \{3,4\}$ are each in one of the boxes 2, 3, and 4, then each of the boxes 2, 3, and 4 has at most two cards. We only need to ensure that box 5 has no more than two cards, i.e., 0 or 1 of $\{2,5\}, \{3,5\}, \{4,5\}$ are in box 5, corresponding to $\mathrm{C}_3^0 + \mathrm{C}_3^1 = 4$ placements.
Therefore, $N = 6 \times 1 + 2 \times 4 = 14$.
By symmetry, there are $4N = 56$ good placements under (2).
In total, there are $64 + 56 = 120$ good placements.