To Prove $ |Y_{n,k}|=(n/k)*{n-2k-1 \choose k-1}$ To prove, $$ |Y_{n,k}|=\frac{n}{k}{n-2k-1 \choose k-1}$$
Where, For $k,n \in  N$, let $Y_{n,k}$ be the collection of $k-element$ subsets $A \subset [n]$ , given that,
$i-j$ is not congruent to 1 or 2 modulo $n$, for all $ i, j \in A$.
Due to the congruence restriction, 
I tried taking a set with 3 subsets within it. Each subset has 3 elements, and I was able to see that formula holds true for that set.
But I can't come up with a methodical way to actually prove this.
 A: Consider the numbers from $1$ to $n$ arranged in a circle. Fix some number $x$, and add it to our set. Then the two numbers at most two to its left and at most two to its right cannot also be picked, leaving $n-5$ numbers we can add to our set. Unlike before, however, these remaining numbers are arranged in a line, so the congruence restriction has been simplified to just saying that we cannot pick two numbers to add to our set that have a distance of at most $2$ on the line. For convenience, we can just relabel our elements as $1, 2,\ldots, n-5$. The picture below demonstrates what I mean. 
Our goal now is to pick a size $k-1$ subset $\{a_1, a_2,\ldots, a_{k-1}\}$, ordered from least to greatest, of the remaining $n-5$ numbers such that $a_{i+1}-a_i\geq 3$ for all $i\leq k-2$. Let $\mathcal{A}$ denote the collection of all such subsets. Our claim is that there is a bijection between $\mathcal{A}$ and the collection of all size $k-1$ subsets of $n-2k-1$. If $\{b_1, b_2, \ldots, b_{k-1}\}$ is a subset of $n-2k-1$, also ordered from least to greatest, then we claim that $\{b_1, b_2+2, b_3+4,\ldots, b_i+2(i-1),\ldots,
b_{k-1}+2(k-2)\}$ is an element of $\mathcal{A}$.
Since $b_{i+1}-b_i\geq 1$, it follows that $b_{i+1}+2i-(b_i+2(i-1))\geq 3$. Since $b_{k-1}\leq n-2k-1$, it follows that $b_{k-1}+2(k-2)\leq n-2k-1+2k-4=n-5$. Conversely, if $\{a_1, a_2,\ldots, a_{k-1}\}\in\mathcal{A}$, then $\{a_1, a_2-2,\ldots, a_i-2(i-1),\ldots, a_{k-1}-2(k-2)\}$ is a size $k-1$ subset of $[n-2k-1]$, and that this is the inverse to the above correspondence, so it is in fact a bijection.
What remains is that there are $n$ ways to choose the original element $x$ which we first added to our set, so we multiply by $n$. However, of the $k$ elements of our set, any one of them could have been chosen to be $x$, so we divide by $k$. This completes our proof.
As an example, let $n=13$, and $k=4$. Say we choose $x=1$. The remaining numbers $4, 5,\ldots, 10, 11$ we relabel as $1, 2,\ldots, 7, 8$. Below, I show the correspondence between the size $3$ subsets of $[8]$ such that any pair of elements differ by at least $3$, and the size $3$ subsets of $[n-2k-1]=[4]$.
$\{1,2,3\}\to \{1, 4, 7\}$
$\{1,2,4\}\to \{1,4,8\}$
$\{1,3,4\}\to \{1,5,8\}$
$\{2,3,4\}\to \{2, 5, 8\}$
