Permutations in Two Rows I have been looking at linear and circular permutations. I have now come across a question that entails permutations in two rows.
This is the question:
Six natives and two foreigners are seated in a compartment of a railway carriage with four seats on either side. In how many ways can the passengers seat themselves if:
i)  the foreigners sit opposite each other
ii) the foreigners do not sit next to each other
My futile attempts:
i) 8! - (7!*2)
After looking at this example, A classroom has two rows of eight seats... 
I tried
8! -(6!/2*4P3)
whose computational value is unfortunately the solution to ii)
My incorrect reasoning in this case was that if 2 people sit opposite each other always, then there are 6 positions left which can be arranged in 2! and for each row you can only choose 3 people to be in any one of the 4 positions.
The correct approach and reasoning for both problems would be greatly appreciated.
 A: For i) you should seat the foreigners first.  Seat one-8 choices.  The other foreigner is fixed. Then seat the natives, $6!$ possibilities.  Total is $8 \cdot 6!=5760$  For ii) again seat one foreigner first.  If he is on the end ($4$ places), there are $6$ left for the other foreigner, then $6!$ for the rest.  Otherwise ($4$ places) there are $5$ for the other foreigner, then $6!$ for the rest.
A: i) Since the condition is imparted on the foreigners, start with them.
Number of ways of choosing a seat for a foreigner $= 4$. If this is done then automatically the seat for the other foreigner will be fixed as he is to be seated exactly opposite to this foreigner. But since there are two rows (opposite to each other), the foreigners could swap their seats with each other still satisfying the condition that they are sitting opposite to each other.Therefore number of arrangements $= 4*2 = 8$.
Now the natives can arrange themselves in the rest $6$ seats in $6!$ ways.
Total possible arrangements $= 4*2*6! = 5760$
ii) Total possible arrangements $= 8!$
Let the two foreigners are adjacent to each other. The number of ways in which the two foreigners can sit adjacent to each other is $2!*3 = 6$ ways on each side, making it a total of $12$ ways. The other $6$ natives can arrange themselves in $6!$ ways.
Total required arrangements $= 8! - 2!*6*6! = 31680$
