Let $H=\{1,2,3,4,5,6,7\}$ and $K=\{1,2,3,4\}$ and let $P$ be the power set of $H$ and let $R$ be the relation on $P$ defined by: For all $X,Y$ in $P$ : $(X,Y)$ are in $R$ if and only if $|X \cap K| = |Y \cap K|$.
how many elements does the equivalence class of $\{1,2\}$ has? I don't know how to solve this one. Can you please help me?
Thanks
For #2, there are $2^3$ total sets (power set of $\{5,6,7\}$) that could work, and you have correctly listed 4 of them.
For #3, a subset in that equivalence class must consist of some subset of $\{1,2,3,4\}$ of size two, combined with some elements outside of $K$. (Effectively, add each subset found in #2 to any subset of $K$ of size $2$.) There are $\binom{4}{2}$ subsets of $K$ of size $2$, and $2^3$ subsets from #2, so in total there are $\binom{4}{2} 2^3= 48$ subsets as you have correctly computed.
Apologies for my earlier mistake.