Countrymen Sitting Around A Table Two Americans, two Canadians, two Mexicans, and two Jamaicans are seated around a round table. Countrymen are distinguishable.
In how many ways can all eight people be seated such that at least two pairs of countrymen are seated together?
Two seatings which are rotations of each other are to be considered the same.
I know this question has been asked earlier but the asker excluded a few important details, such as that rotations are counted as identical and that it was at least two pairs of countrymen sitting together.
I know I have to use PIE(Principle of Inclusion and Exclusion) in order to solve this problem, but I have no idea how.
 A: We break symmetry by assuming one of the chairs is a throne, and the American Trumperor sits on the throne.
We first count the number of ways to seat the people so that Americans are together, and Canadians are together. The remaining American can be seated in $2$ ways. Tie the two Canadians together. The $4$ remaining people and the Canadian bundle can be arranged in $5!$ ways, and then the Canadians can be unbundled in $2$ ways,  for a total of $5!2^2$.
The same is true for Americans together and Mexicans together, and Americans together and Jamaicans together.
Out of caution we now count Canadians together and Mexicans together. That is two bundles and $3$ others (the two Jamaicans and the remaining American), and these can be arranged in $5!2^2$ ways. 
We obtain a total of $\binom{4}{2}5!2^2$.
But we have multiple-counted the arrangements in which the Trumperor sits on the throne, and there are three groups of the same nationality together. 
Each choice of group of $3$ has been counted $3$ times, and should only have been counted once.
A count similar to the one above shows that there are $\binom{4}{3}4!2^3$ such arrangements. So we adjust by adding $2\binom{4}{3}4!2^3$. 
Thus our revised estimate is $\binom{4}{2}5!2^2-2\binom{4}{3}4!2^3$.
But we have subtracted too much, we must add back the arrangements in which all nationalities are together (and the Trumperor sits on the throne). There are $\binom{4}{4}3!2^4$ such arrangements. Each has been subtracted $4$ times, and should only have been subtracted once.
Our final count is $\binom{4}{2}5!2^2-2\binom{4}{3}4!2^3+3\binom{4}{4}3!2^4$.
A: This approach, too, consists of breaking up into disjoint cases, and adding up.
I shall, however, treat the exactly $2$ couples together case differently
I shan't dwell much on the two easier cases. 


*

*All $4$ couples together: $3!\cdot 2^4 = 96$

*Exactly $3$ couples together: $\binom43\cdot2!\cdot 2^3 \cdot3\cdot2 = 384$

*Exactly $2$ couples together: $\binom42\cdot2^2 = 24$ ways for seating two couples.
Name these couples $X, Y,$ and the two that will provide "singles", $A,B,$
and "straighten" the circle, $\;\;Y\quad X\quad Y$
Either both the $A's$ will be on one side (with a $B$ sandwiched in) or on both sides of $X$, as under:
$Y - ABA-X-Y:\;\; 3\cdot2\cdot 2^2 = 24\;\;$ (places for $2_{nd}B)\times$(sides for $A)\times$(perms of $A$ and $B$)
$Y-A-X-A-Y:\;\; 2\cdot (4\cdot3) = 24\;\;$ (perms of $A)\times$(place $B's$)
thus $24(24+24) = 1152$  
Adding up, $96+384+1152 = 1632$
A: You don’t actually need the PIE: you can count the arrangements with $2,3$, and $4$ pairs seated together.
Suppose that all four pairs are seated together. There are $3!=6$ cyclic orders for the pairs, and each pair can be seated in either of $2$ ways, so we get a total of $6\cdot2^4=96$ arrangements.
If exactly three pairs are seated together, there are $4$ ways to pick the three pairs, $2!=2$ cyclic arrangements of the pairs, and $2$ ways to seat each pair. The remaining two people can be inserted in any $2$ of the $3$ slots between pairs; there are $\binom32=3$ ways to choose the slots and $2$ ways to fill them, so the final total for this case is $4\cdot2\cdot2^3\cdot3\cdot2=384$ arrangements.
The only really messy case is the last one: exactly two pairs seated together. There are $\binom42=6$ ways to choose the two pairs and $2$ ways to seat each pair, so we start with a factor of $6\cdot2^2=24$. Now we have to count the ways to insert the remaining $4$ people into the $2$ slots between the pairs. 


*

*If we put $3$ in one slot and $1$ in the other, there are $4$ ways to choose the singleton and $2$ ways to choose his slot, and $2$ ways to insert the other $3$ people into the other slot: his countryman must be between the other two, and the only choice is which way around they sit. For this subcase, then, we have a factor of $4\cdot2\cdot2=16$.  

*If we split them $2$ and $2$, we must pair each of them with a non-countryman; there are $2$ ways to do this. There are then $2$ ways to assign these mismatched pairs to the $2$ slots, and each mismatched pair can be seated in $2$ different orders, so this subcase also gives us a factor of $2^4=16$.  

*If we put all $4$ in one slot, there are $2$ ways to choose the slot. The four people must alternate countries within that slot. If the countries are $X$ and $Y$, the alternation clockwise can be $XYXY$ or $YXYX$; that’s a factor of $2$. Each pair of countrymen can appear in either order; that’s two more factors of $2$. Altogether we get a factor of $2^4=16$ again.


The total for this case is therefore $24(16+16+16)=1152$, and the grand total is $1632$.
