In how many ways is it possible to seat seven people at a round table if Alex and Bob must not seat in adjacent seats? I am SUPER CONFUSED about circular permutations can anyone help me with the following problem.
Here is one example from the AMC 10:
In how many ways is it possible to seat seven people at a round table if Alex and Bob must not seat in adjacent seats?
 A: There's a general trick for solving many of these kinds of constrained arrangement problems: often it's easier to count the number of arrangements that don't satisfy the constraint, and then subtract this from the total number of possible unconstrained arrangements.
So, start with the total number ways to arrange seven people around a circular table.  Then subtract from this the number of arrangements where Alex and Bob are sitting next to each other.
So, how to count those, then?  Well, if Alex and Bob must always sit next to each other, then we can pretend that they're really just one person.  Then we have six people (where one of those "people" is really Alex + Bob) that we can freely arrange around the table.  And we can obviously count those arrangements exactly the same way as we counted the number of ways to arrange seven people around the table.
…but don't forget that there are two ways to seat Alex and Bob next to each other: either Alex can sit on the left and Bob on the right, or vice versa.  So we actually have to subtract twice the number of ways to arrange six people around a circular table from the number of ways to arrange seven people around the table to get the final answer.
A: You must know that the general formula for seating at a  round table is $\frac{n!}{n}$
So seat the people excluding Alex and Bob in $\frac{5!}{5}$ ways
Now there will be $5$ gaps between every two people. Seat Alex and Bpb in the gaps in $5\cdot4$ ways.
Finally multiply to get $\frac{5!}{5}\cdot5\cdot4$ ways
