How to derive a general formula for this problem? (pairs of people seated around a table) 
$N$ people attend a dinner party and sit round a circular table.Each
  person knows only the people sitting immediately next to him and has
  to be introduced to everyone else.If the total number of pairs of
  people introduced to each other is $20$,then what is the value of $N$?

Is this a form of any well known problem?I could only of think of a brute-force solution,however I am inquisitive to know how to solve this for any number of pairs in an efficient manner.
ADDED:It seems like I have understood the pattern,for any $N$ the number of pairs of introduction is given by $(N-3)+(N-3)+(N-3)-1+(N-3)-2+\cdots+2+1$ by using this idea the solution of the above problem should be $N=8$.
 A: Each person is introduced to $N-3$ others.  As there are $N$ people, that would make $N(N-3)$ introductions, but each has been counted twice-once from each side.  So we have 20=$\frac{N(N-3)}{2}$, which has roots $8, -5$.  
A: Let $x$ be the number of introductions.
Suppose that everyone must be introduced to everyone else.  Do this as follows.
(i) Everyone is introduced to her left neighbour ($N$ introductions);
(ii) Every pair of non-neighbours is introduced ($x$ introductions).
It follows that
$$\binom{N}{2}=N+x,\qquad\text{and therefore}\qquad x=\binom{N}{2}-N=\frac{N^2-3N}{2}.$$ 
Now we solve the equation
$$\frac{N^2-3N}{2}=20.$$
The only admissible solution is $N=8$.
A: $nC2 - n = 20$
$n(n - 1)/2 - n = 20$
$n(n - 1) - 2n = 40$
$n^2 - 3n - 40 = 0$
$n^2 - 8n + 5n - 40 = 0$
$n(n - 8) + 5(n - 8) = 0$
$(n - 8)(n + 5) = 0$
$n = 8$ or $n = -5$
now n is positive and greater than $3$ ($3$ people needs no introduction) , so we take $n = 8$
A: Total introductions needed(if nobody knows anyone)=$n\choose 2$; but since a person knows the person next to him, therefore introductions needed=${n\choose 2}-n$ which is given to be $20\implies \frac{n(n-1)}{2} - n=20\implies n^2-3n-40=0\implies n=8$.
