Countrymen seated around a round table probability question Seated around the table are:
- 2 Americans
- 2 Canadians
- 2 Mexicans
- 2 Jamaicans. 
Each countryman is distinguishable. 
How many ways possible can all 8 people be seated such that AT LEAST TWO men from the same country sit next to each other? 
Rotations are considered the same.
I know that I have to use PIE somehow on the two men, but it is slightly more confusing because there are four groups and two men in each group. Any hints?
EDIT 12/1/2014: I checked the original problem statement and it said that "In how many ways can all eight people be seated such that at least two pairs of countrymen are seated together?". I mistakenly put "at least two men" rather than "at least two pairs". I'm terribly sorry for the confusion!

Let us try to find those arrangements in which no two countrymen sit together.Let A1A2,B1B2,C1C2 and D1D2 represent same countrymen. Since we are talking about circular permutations,let A1 be reference point and let B1,C1 and D1 be seated at poition 3,5 and 7 respectively. 
              1 
              A1

      8               2


   7 D1               B1 3


      6               4


              C1
              5

At position 2 we can have C2,D2 only,at 4-A2,D2 at 6-A2,B2 and at 8-B2,C2.If C2 is chosen at 2, other places get fixed since 8 can have B2 only,6-A2 and 4 D2. Thus, we can have only 2 permutaions if we select 2 as C2 and D2. 
Now,since postions 3,5 and 7 may be selected with any 3 people out of 7 thus we have a total of P(7,3)*2 permutations in which no two sit together.
Thus 7!-P(7,3)*2 permutations are those in which two or more countrymen sit together.
