A ten-member group went to the cinema. They got tickets for two different rows, with 5 seats each next to each other. From the group, Ábel and Bendegúz would like to sit next to each other, while Zsuzsi and Anikó want to sit separately. In how many ways can they be seated?
I. solution. If we calculate how many cases there are in which Ábel and Bendegúz sit next to each other, and subtract from this number the number of possibilities where, in addition to them, Zsuzsi and Anikó also sit next to each other, we will get the answer to the question posed in the problem.

First, let's calculate the number of cases in which Ábel and Bendegúz sit next to each other. Consider them as one person. In this case, 4 people sit in one row, and 5 in the other. There are 2 possibilities for which row has 4 people. For which 3 people sit with Ábel and Bendegúz, there are $\binom{8}{3}=56$ possibilities. The number of possible arrangements in each row is 4! and 5!, and Ábel and Bendegúz can also switch places with each other.

This gives $2 \cdot 56 \cdot 4! \cdot 5! \cdot 2 = 645120$ possibilities.

Now let's calculate the number of cases in which, in addition to Ábel and Bendegúz, Zsuzsi and Anikó also sit next to each other. Consider them as one person as well.

If the two pairs sit in the same row, there are $\binom{6}{1}=6$ possibilities for choosing the person sitting next to them. The two pairs and this person can sit in the row in $3! \cdot 2 \cdot 2 = 24$ ways, since the members of both pairs can switch places with each other. The 5 people in the other row can sit in 5! ways. Finally, which row the pairs sit in can be chosen in 2 ways.

This gives $6 \cdot 24 \cdot 5! \cdot 2 = 34560$ possibilities.

If the two pairs sit in different rows, there are 2 ways to choose Ábel and Bendegúz's row. For choosing 3 people to sit with Ábel and Bendegúz, there are $\binom{6}{3}=20$ possibilities. In both rows, the number of possible arrangements for the 3 people and the pair is 4!. The members of both pairs can switch places with each other.

This gives $2 \cdot 20 \cdot 4! \cdot 4! \cdot 2 \cdot 2 = 92160$ possibilities.

The number of cases in which, in addition to Ábel and Bendegúz, Zsuzsi and Anikó also sit next to each other is $34560 + 92160 = 126720$. Therefore,

$$
645120 - 126720 = 518400
$$

is the number of possible arrangements.

II. solution. We break the problem into three cases.

Case I: Ábel, Bendegúz, Zsuzsi, and Anikó sit in the same row. In this row, Ábel and Bendegúz can sit at the end of the row, and the number of possibilities for this is $2 \cdot 2 \cdot 2 \cdot 6 = 48$, since the row has two ends, Ábel and Bendegúz can switch places, Zsuzsi and Anikó can occupy the remaining three adjacent seats in two ways, and one of the remaining six people can sit in the last seat. If Ábel and Bendegúz do not sit at the end of the row, the number of possibilities is similarly $2 \cdot 2 \cdot 4 \cdot 6 = 96$.

In the other row, the remaining people can be arranged in 5! ways.

In Case I, therefore, $2 \cdot (48 \cdot 5! + 96 \cdot 5!) = 34560$ seating arrangements are possible, since Ábel and Bendegúz can sit in either row.

Case II: Zsuzsi and Anikó sit in one row, but not in the same row as Ábel and Bendegúz. In this case, Ábel and Bendegúz can sit in $4 \cdot 2 = 8$ ways, and three people can sit next to them in $6 \cdot 5 \cdot 4 = 120$ ways, which is $8 \cdot 120 = 960$ possibilities. In the other row, Zsuzsi and Anikó can sit in $3 + 2 + 2 + 2 + 3 = 12$ ways, and the remaining 3 people in 6 ways, which is 72 possibilities.

In Case II, therefore, $2 \cdot (960 \cdot 72) = 138240$ seating arrangements are possible.

Case III: Exactly one of Zsuzsi and Anikó sits in the same row as Ábel and Bendegúz. There are 2 possibilities for which one it is. In this row, Ábel and Bendegúz can sit in 4 places, they can switch places, and Zsuzsi or Anikó can sit in 3 places, while the other 6 people can sit in $6 \cdot 5 = 30$ ways. This gives $2 \cdot 4 \cdot 2 \cdot 3 \cdot 6 \cdot 5 = 1440$ possibilities. In the other row, the remaining people can sit in $5! = 120$ ways.

In Case III, therefore, $2 \cdot (1440 \cdot 120) = 345600$ seating arrangements are possible.

In total, $34560 + 138240 + 345600 = 518400$ seating arrangements are possible.