[Math] Circular Permutations problem

Question

In how many ways can two Americans, two British, one Chinese, one Dutch and one Egyptian sit at a round table so that persons of the same nationality are always separated? (Clockwise and anti-clockwise arrangements are different.)

My attempt:
I first arranged the Dutch, Egyptian and Chinese gentlemen on the round table in such a way that there is one empty chair between any two persons and two empty chairs between a single pair. This can be done in 2! ways. Then I choose a Britishman and an American each as well as two seats for them and arrange them in 2! ways. Now the remaining pair of an American and a Britishman have to sit together and the no. of internal arrangements for then would also be equal to 2!. Multiplying all of these gives me answer 96. But my book mentions the answer as 336. I think my method is incorrect. Can anyone suggest me the correct method?

Answer 1

Fix the Egyptian. Then we can use the inclusion–exclusion principle to find the number of arrangements:

• There are $6!=720$ ways to seat the other people, ignoring restrictions.
• Of these, $5!\cdot2=240$ ways seat the Americans together and are excluded (this may be seen by treating the Americans as a block; note that they can sit in two ways internally). The same number applies for the British.
• $4!\cdot2\cdot2=96$ ways seat both Americans and both British together, and are included.

So there are $720-240-240+96=336$ ways to seat the people with the given restrictions.