Let $S = \{1,2,3,4,5,6,7,8,9\}$ and let $T = \{2,4,6,8\}$. Let $R$ be the relation on $\mathcal{P}(S)$ defined by $\forall X, Y \in \mathcal{P}(S), (X,Y) \in R$ iff $|X - T| = |Y - T|$.
How many elements of $[\{1,2,3,4\}]$, the equivalence class of $\{1,2,3,4\}$, are there?
If $[\{1,2,3,4\}] = 4$, then using combinations, I could get an answer of $80$ elements.
But if $[\{1,2,3,4\}] = {5 \choose 2} * | \mathcal{P}(T) |$, then I get $\frac{5!}{2!3!} * 2^4 = 160$ elements.
The number of subsets $X$ of $S$ such that $|X \setminus T| = a$ is $\binom{\overline{T}}{a} 2^{|T|} = 16 \cdot \binom{5}{a}$: we need to choose $a$ elements out of $\overline{T}$, and any number of elements from $T$. In particular, the equivalence classes have the following sizes: $$ \begin{array}{c|cccccc} a & 0 & 1 & 2 & 3 & 4 & 5 \\\hline \text{size} & 16 & 80 & 160 & 160 & 80 & 16 \end{array} $$ Your set has $a = 2$, so the equivalence class contains 160 subsets.