Find the number of ways of sitting in the stated order on circular table 
In how many ways can $5$ Americans and $6$ Indians sit across a circular table such that no two Americans are sitting next to each other?

$6$ Indians can sit in $(n-1)!$ ways on a circular table or $5!$ ways.
So for $5$ Americans there are $6$ positions which it can take, Now my doubt is that since there are $5$ Americans and $6$ positions which it can take, will the no of ways be $5!$ or $6!$ ?
If so then why?
 A: Unless otherwise specified, people are taken to be distinct.
Considering chairs to be unnumbered, $6$ Indians can be seated in $6$ chairs in $5!$ ways
There are 6 interstices between the Indians where chairs can be pushed in for the Americans, so the 1st has 6 choices, the 2nd 5 choices, and so on.
Finally, # of arrangements = $5!*6!$
[ You can note in passing that $^6P_5$ =$^6P_6$ ]
A: First of all, no one wants to sit with Americans! joking!!! 
Well, you need to realize that the number of Indians has to be at least the same amount as Americans at a round table. (at most one less than them if it's not a round table.)
Now, if there are only Indians sitting at the table, how many spots are left for Americans? Given no two Americans can sit together, we then have 6 spots left for 5 Americans. So they can sit anywhere in those 6 spots. Therefore, the answer is ${6\choose 5}$ which gives you the answer 6. Assuming there is not difference between any two Americans, and no difference between any two Indians. 
Another thing you need to check is that, since they are sitting at a round table, is this empty seat A, for example, the same as the empty seat B? What I'm saying is that, for example if you have a 2-seat round table, one person sitting at seat A leaving seat B empty vs sitting at B leaving A empty. Are they counted as the same arrangement at this round table or not? If they are counted as the same then you only have one arrangement~~
