Probability question?

For convenience, I use "a", "b" and "f" to represent American, British and French, respectively.

1)
The order of seats is considered.
The number of all outcomes (the order of seats considered) = 30 * 29 * 28 * 27 * 26 = 17100720.
2 seats, and 5a, 9b and 13f remain, so
There are 3 combinations: a X b, a X f, and b X f, and so
5 * 9 + 5 * 13 + 9 * 13 = 227 outcomes of the remaining people.
Multiply 227 by 2 as
The remaining people, who get the remaining 2 seats, can interchange them.
Then multiply 454 by 2 as
The 3rd and 4th French people can interchange their seats, too.

Therefore, the probability is 908/17100720 = 227/4275180.

2)
The order of seats is considered.
The number of all outcomes (the order of seats considered) = 30 * 29 * 28 * 27 * 26 = 17100720.
If A and B are picked, then 3 seats and 28 people remain. Hence we get
28 * 27 * 26 = 19656 outcomes of the remaining seats and people.
Multiply 19656 by 5 * 4 as
There are 20 outcomes that either A or B has 5 choices of the seats, and then B (or A) has 4.

Therefore, the probability is 393120/17100720 = 2/87.