number of ways in which $4$ people can be chosen from $16$ people Total  number of ways in which $4$ people can be chosen from $16$ people sitting in a circle such that no two of them are neighbors.
Number of ways of choosing $4$ persons out of  $16$ persons, is $\displaystyle \binom{16}{4}$.
Won't be able to go further, could someone help me with this, thanks.
 A: The number of ways to choose $k$ non-consecutive 
persons on a circle of size $n$ is -
${n-k+1 \choose k} - {n-k-1 \choose k-2}$
First term is the number of ways to choose k non-consecutive positions on a size n and the second term subtracts off those arrangements where positions 1 and n were both chosen.
I am taking reference Aryabhata explanation in the answer here.
A: Without any restrictions, there are $\dbinom{16}{4}$ ways
$16$ ways where all four people are neighbors
$16\dbinom{11}{1}$ ways where exactly three people are neighbors
Now for the case when  two people are neighbors, no. of ways would be
$$16\dbinom{12}{2}-\frac{16*11}{2}$$
$16\dbinom{12}{2}$ might look as the number of ways in which $2$ people are also sitting together. But we are counting twice the cases in which $2$ people are neighbors and the other $2$ are also sitting together.
So, first we have to count the number of cases in which $2$ pairs of people are sitting together.
There would be $16$ ways of selecting $2$ people sitting together.
Now for the other two we are left with $12$ people to choose from. There would be $11$ ways to select other $2$ neighboring people. Right now we have $16*11$ but we double counted each pair of neighbors. Ask me in the comments if you didn't get this part.
So, it would be $16*11/2$.
Thus the answer is
$$\dbinom{16}{4}-16-\left(16\dbinom{12}{2}-\frac{16*11}{2}\right)-16\dbinom{11}{1}=660$$
