Circular permutations problem, need help 
How many ways can $5$ people sit on a circular table of $8$ places?

I know that if both people and  places number =$n$ , then the answer $= (n-1)!$
But the situation here is different, since there are more places than people,  what should I do?  
Thanks for help

Let $X$ be the set of all possible arrangements of $5$ people in a straight line of $8$ chairs. Now, define a relation on $X$ : two arrangements are similar, if they are related by a cyclic permutation: if there exists a number $0 \leq r\leq 7 $, such that each person in the first arrangement, when shifted by $r$ places to the right, lands up where he is supposed to be in the second arrangement (with wrapping around i.e. from extreme right the next shift is to the extreme left).
Now, note that this relation is an equivalence relation, and the equivalence class of each straight line permutation contains eight elements, and represents one single arrangement around a circular table.
Therefore, the answer to your question is just the number of ways that $5$ people can sit in $8$ places in a straight line(that is, with no constraints on symmetry), divided by $8$. The answer to this question is then just $\frac{8 \times7\times6\times 5\times 4}{8} = 7\times6\times5\times4 = 840$.
