A group of $n\geq 6$ members decide to split up and travel in $n-3$ parties. How many equivalence relations exist on the set of members such that the members in each travel party form an equivalence class?
I'm finding equivalence relations to be challenging in my discrete math course and I'm expected to know the material for an upcoming midterm so could someone help guide me through a solution for the above problem?
Think about the size of the $n-3$ parties. For concreteness, take $n=10$. We want $7$ groups out of $10$ people. Clearly most people will be in a group by themselves. We could have $10=1+1+1+1+1+1+4$, which has six singleton groups and one group of four. If this is the group division, then there are ${10\choose 4}=210$ ways to pick the group of four; after this is selected, the groups are all determined. These are $210$ possible equivalence relations which are a partial answer to the question.
We could also have $10=1+1+1+1+1+2+3$, which has five singleton groups, one pair, and one group of three. There are ${10\choose 3}=120$ ways to pick the group of size $3$, and ${7\choose 2}=21$ ways to pick the pair from the remaining people. Hence there are $120\cdot 21=2520$ possible equivalence relations for this scenario.
There's also $10=1+1+1+1+2+2+2$, which I will leave for you to consider. It turns out that there are always just three possibilities, for all $n\ge 6$ (not just $n=10$).