Choosing seats for guests You have a circular table with $N$ seats.$K$ bellicose guests are going visit your house of-course you don't want them to sit beside each other.As the host, you want to find out how many ways there are to choose $K$ seats such that none of them is adjacent to each other. 
I noticed that there is a solution  other than $0$ if ($N \ge 2K $) but I am not sure how to approach for the rest.
EDIT: Only $K$ bellicose guests are visiting,no friendly guest are there the remaining $N-K$ seats will be vacant.
A possible mathematical translation of this problem: Choosing $K$ candidate points from a circle of $N$ indistinguishable points such that there are more than one vacant point between adjacent candidate points.
 A: Choose a seat $S$, and a bellicose guest $B$. Sit $B$ in $S$, and tell them not to move, whether they like it or not. That done, there are $(K-1)!$ ways of ordering the remaining bellicose guests clockwise around the table, and $F!$ ways of ordering the friendly guests (here I am using leonbloy's notation $F = N - K$). For each such ordering, we have to choose a pattern of the form $f...b...f...b...f$. This pattern:
1. starts and ends with $f$ (so that $B$ is isolated);
2. contains $(K-1)$ $b$'s and $F$ $f$'s; and
3. contains no two adjacent $b$'s.  
But the number of such patterns is the same as the number of patterns that
1. start with $f$; and
2. contain $(K-1)$ $b$'s and $(F-K+1)$ $f$'s.  
(To see this, just replace each instance of $bf$ in the original pattern by $b$.) The number of such patterns is the binomial coefficient $\binom{F-1}{K-1}$. So we end up with:  
$(K-1)!F!\binom{F-1}{K-1} = \frac{F!(F-1)!}{(F-K)!}$  
This is the number of seating arrangements with guest $B$ in seat $S$. Multiply by $N$ to get the total number.
Edit Reading the question more carefully, it asks for the number of (what I call here) patterns, not the number of seatings. For each pattern, there are $K!F!$ seatings, so the answer is  
$N\frac{F!(F-1)!}{(F-K)!}/(K!F!) = \frac{N(F-1)!}{(F-K)!K!}$
A: Let's call $F=N-K$ number of "friendly" guests. And let's call $S(F,K)$ the count of seating ways assuming that seats and guests are distinguible (rotations are distinct solutions).
We know that $F \ge K$. For the limit case $F = K$ we have $S(F,K) = 2 K (K-1)! K! = 2 K!^2$.
Now, the recursion: we count the number of ways when adding a friendly guest:
$S(F+1,K) = (F+K+1) S(F,K)$ 
From this (if it's correct!) you can get an explicit solution. 
Update: this is not correct. See TonyK's answer
