Permutation in line and circle with conditions. A party of $10$ consists of $2$ Americans, $2$ British men, $2$ Chinese &amp; $4$ men of other nationalities (all different). Find the number of ways in which they can stand in a row so that no two men of the same nationality are next to one another. Find also the number of ways in which they can sit at a round table.
Here for the first one I try to use the gap method. So arrange $5$ to create gaps to accommodate all but we have only four of different nationalities. So a different method should be used.
All permutations - permutations of people of same country being together
But this has cases as only one pair together, two pairs together and three pairs together. Which overlap. How to handle ?
In a row:  There are $10!$ ways to arrange the people. We now count the bad arrangements, in which two A's are next to each other, or two B's, or two C's. 

To do the counting, we use Inclusion/Exclusion. How many arrangements have the two A's together? Put them in a bag. We now have $8$ people and $1$ bag. These can be arranged in $9!$ ways. Then when we let the A's out of the bag, they can arrange themselves in $2$ ways, for a total of $2\cdot 9!$.

Do the same for the B's, the C's, and add up. We get $3\cdot 2\cdot 9!$. 

However, we have double-counted the bad arrangements in which, for example, the A's and the B's are together. The same bag argument shows there are $2^2\cdot 8!$ such arrangements, for a total of $3\cdot 2^2\cdot 8!$.

Thus our next estimate for the number of bads is $3\cdot 2\cdot 9!-3\cdot 2^2\cdot 8!$.

However, we have subtracted one too many times the arrangements in which the A's are together, and the B's, and the C's. The now familiar technique  shows there are $2^3\cdot 7!$ of these. So add back $2^3\cdot 7!$.

We end up with $7!(6\cdot 9\cdot 8-12\cdot 8+8)$, which is $8!\cdot 43$. 

This is the number of bad arrangements. The number of good arrangements is $10!-8!\cdot 43$, which is $8!\cdot 47$.

Circular table: We use the convention that two orders are to be considered the same if they differ by a rotation. Assume that one of the "others" is a Canadian. He/she is probably too polite to complain, so one can put the Canadian in the worst chair.

The remaining chairs can now be thought of as a line. We have $9$ people, including $3$ "others" to put in a line, with the condition that the two A's, two B's, and two C's are separated. Same Inclusion/Exclusion technique.
For the first one, Total no of combinations is, $$ s = \frac{ 10!}{2!.2!.2!} $$ Now let A represent the group in which two Americans are together, B represent 2 British together and C represent group in which 2 Chinese are together, calculate $  A \cup B \cup C$ and subtract it from s.

So the final answer is $$ \frac{ 10!}{2!.2!.2!} - 3.\frac{9!}{2!.2!} + 3.\frac{8!}{2!} - 7! $$