Relations of equivalence... Given the set $A = \{1,2,3,4,5,6,7\}$. 
How can I establish how many and what are the equivalence relations $R$ on $A$ that meet all the following conditions?
1) $|A/R| = 3$
2) $1 R 2$
3) $[2]_R\neq[3]_R$
4) $[4]_R = [5]_R$
5) $|[7]_R| = 3$
 A: Since partition and equivalence relation are equivalent, it suffices to consider the ways to partition $A$.


*

*$|A/{\cal R}| = 3$: there are three partitions

*$1 {\cal R} 2$: $1$ and $2$ are in the same partition

*$ [2]_{\cal R} \ne [3]_{\cal R}$: $2$ and $3$ are in different partitions

*$[4]_{\cal R} = [5]_{\cal R}$: $4$ and $5$ are in the same partition

*$|[7]_{\cal R}|=3$: partition $[7]_{\cal R}$ has three elements.


We start with $\{\{1,2\},\{3\},\{\}\}$.


*

*$[4]_{\cal R} = [1]_{\cal R}$
This gives $\{\{1,2,4,5\},\{3\},\{\}\}$.  Wherever we put $7$, property (5) is violated.

*$[4]_{\cal R} = [3]_{\cal R}$
From $\{\{1,2\},\{3,4,5\},\{\}\}$ and property (5), the only choice is $\{\{1,2,7\},\{3,4,5\},\{6\}\}$.

*Otherwise, $\{\{1,2\},\{3\},\{4,5\}\}$.  Use property (5) again.


*

*When $7$ is in the first or the third partition, $6$ can be in either of the other two partitions.

*When $7$ is in the second partition, then so as $6$.



We have $1+2\times2+1=6$ choices.
\begin{align}
& \{\{1,2,7\},\{3,4,5\},\{6\}\} \\
& \{\{1,2,7\},\{3,6\},\{4,5\}\} \\
& \{\{1,2,7\},\{3\},\{4,5,6\}\} \\
& \{\{1,2\},\{3,6,7\},\{4,5\}\} \\
& \{\{1,2,6\},\{3\},\{4,5,7\}\} \\
& \{\{1,2\},\{3,6\},\{4,5,7\}\} \\
\end{align}
