Selecting $k$ persons from round table, no two of whom are adjacent Consider round table with $n$ persons, we want to choose $k$ persons from them, no two of whom sit with each other. So there are no neighbours. 
Lets enumerate them : $(a_1 \dots a_n)$, but lets consider $(a_1 a_2 \dots a_n a_1)$
My attempt.
There are two cases :
1) $10 a_3\dots a_{n-1} 01$ the case when we select first one. So we need to add $01$ in row of series. There is a $\binom{n-k}{k-1}$ ways to do it.
2) $0 a_2 \dots a_n 0$. Now we need to add $10$ there will be $\binom{n-k-1}{k}$ ways to pick it.
Am I right?
 A: No, you are wrong. For example if $k=1$ according to your formula we find
$$\binom{n-k}{k-1}+\binom{n-k-1}{k}=1+n-2=n-1$$
instead of $n$.
Recall that the number of ways to choose $K$ non-consecutive persons in a row of $N$ persons is $\binom{N-K+1}{K}$ (see  Choosing numbers without consecutive numbers.)
Following your approach, in the first case  we need to choose $K=k−1$ non-consecutive persons from $N=n−3$ in a row: the number of ways is
$$\binom{N-K+1}{K}=\binom{(n-3)-(k-1)+1}{k-1}=\binom{n-k-1}{k-1}.$$
In the second case  we need to choose $K=k$ non-consecutive persons from $N=n−1$ in a row: the number of ways is
$$\binom{N-K+1}{K}=\binom{n-1-k+1}{k}=\binom{n-k}{k}.$$
Hence the total number of ways is
$$\binom{n-k-1}{k-1}+\binom{n-k}{k}.$$
A: The first case comes to finding the number of sums $b_1+\cdots+b_k=n-k$ where the $b_i$ are positive integers, and a $b_i$ can be identified as number of non-chosen persons between two persons who are chosen. 
Setting $c_i=b_i-1$ it comes to finding the number os sums $c_1+\cdots+c_k=n-2k$ where the $c_i$ are nonnegative integers.
With stars and bars we find $\binom{n-2k+k-1}{k-1}=\binom{n-k-1}{k-1}$ possibilities.
The second case comes to finding the number of sums $b_0+b_1+\cdots+b_{k-1}+b_k=n-k-1$ where $b_0, b_k$ are nonnegative integers and $b_1,\dots,b_{k-1}$ are positive integers. Here $b_0$ can be interpreted as the number of non-chosen persons between the (non-chosen) person $a_1$ and the first chosen person on the right side of person $a_1$. For $b_k$ similar but then at the left side.
Setting $c_0=b_0$, $c_k=b_k$ and $c_i=b_i-1$ for $i=1,\dots,k-1$ we must now find the number of sums $c_0+c_1+\cdots+c_k=n-2k$ where every $c_i$ is a nonnegative integer.
With stars and bars we find $\binom{n-2k+k}{k}=\binom{n-k}{k}$ possibilities for this.
