Selecting nonadjacent people in a circle Suppose 19 people are sitting at a circular table.  In how many ways can we select 5 of the people so that no two of them were sitting next to each other?
(My idea is to first consider the situation where they are sitting in a row, but I'm not sure how to adjust this for the circular case.)
EDIT: I am not identifying two rotations of the same seating as being identical.
 A: Your idea is good.
First we count how many subsets of  $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19\}$ of size $5$ don't have consecutive elements.
The answer is $\binom{15}{5}$.
We now remove all the subsets that contain $1$ and $19$ from the count, these are clearly equal to the number of subsets of size $3$ of $\{3,4,5,6,7,8,9,10,11,12,13,14,15,16,17\}$ without two consecutive elements, and there are $\binom{13}{3}$ of those.
$$$$
So the final answer is $\binom{15}{5}-\binom{13}{3}=2717$.
As can be verified by the following c++ code:
#include <bits/stdc++.h>
using namespace std;

int A[20];

int push(){
    for(int i=0;i<19;i++){
        if(A[i]==0){
            A[i]=1;
            for(int j=0;j<i;j++){
                A[j]=0;
            }
            return(1);
        }
    }
    return(0);
}

int check(){
    int tot=A[18];
    if(A[0] && A[18]) return(0);
    for(int i=0;i+1<19;i++){
        tot+=A[i];
        if(A[i]&&A[i+1]) return(0);
    }
    if(tot!=5) return(0);
    return(1);
}

int main(){
    int res=0;
    while(push() ){
        res+=check();
    }
    printf("%d\n",res);
}

A: Let $N$ be the number of ways to choose $5$ "separated" people.
We first count the number of ways to choose $5$ separated people, and designate one as special. By a "line" argument there are $\binom{13}{4}$ ways to designate Alicia as special, and choose $4$ people to join her, so that all are separated.
So there are $19\binom{13}{4}$ ways to designate someone as special and choose $4$ people to join her.
We have $19\binom{13}{4}=5N$, so $N=\frac{19}{5}\binom{13}{4}$.
A: I take it that the seats are numbered, and find it simpler to compute directly for a circle.
Denoting the chosen/unchosen by white/black circles, form $5$ blocks like $\boxed {\large\circ\bullet}$
Now there are $5$ such blocks $+ 9$ others, i.e. a total of $14$ entities
The blocks can be placed among the $14$ entities in $\binom{14}5$ ways,
but you are allowing only $14$ starting positions instead of the legitimate $19,$
thus number of ways $= \dfrac{19}{14}\dbinom{14}{5} = 2717$ 
