We have elements on circle $A = \{1,2,\dots,n\}$ are written in ascending order. Find number of subsets consist of $k$ elements, where every two elements are not neighbours on circle where $ 1 \leq k \leq [\frac{n}{2}]$.
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Let $N_1$ be the number of integer solutions of $$x_1+x_2+\dots +x_k=n-k$$ with $x_i\geq 1$ for $i=1,\dots,k$. By using Stars and Bars technique, we get that $$N_1=\binom{(n-k)-k+k-1}{k-1}=\binom{n-k-1}{k-1}.$$ Note that $N_1$ is number of subsets which consist of $k$ elements AND CONTAIN $1$, where every two elements are not neighbours on circle (consider $x_i$ to be the number of elements between the $i$th element of the subset and $(i+1)$th element).
It remains to evaluate $N_0$, that is the number of subsets which DO NOT CONTAIN $1$. This is the number of integer solutions of $$x_1+x_2+\dots +x_{k+1}=n-1-k$$ with $x_i\geq 1$ for $i=2,\dots,k$ and $x_1,x_{k+1}\geq 0$. Hence $$N_0=\binom{(n-1-k)-(k-1)+k}{k}=\binom{n-k}{k}.$$ Hence the total number of subsets is $$N_1+N_0=\binom{n-k-1}{k-1}+\binom{n-k}{k}=\frac{n}{k}\binom{n-k-1}{ k-1}.$$
P.S. See OEIS's A034807: the number of necklaces of $k$ black and $n - k$ white beads with no adjacent black beads.
Robert Z
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Why sum of differences is equals to $n-k$ ? – duncan Sep 17 '17 at 13:12
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What is $x_k$ ??? we have got only $k$ elements in subset – duncan Sep 17 '17 at 13:25
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$x_k$ is the number of elements between $k$th element of the subset and the FIRST one (we are on circle). – Robert Z Sep 17 '17 at 13:27
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May be you can help with other one https://math.stackexchange.com/questions/2434965/find-the-number-of-n-husbands-placing – duncan Sep 18 '17 at 19:16