Guy with $7$ friends Some guy has $7$ friends$(A,B,...,G)$. He's making dinner for $3$ of them every day for one week. For how many ways can he invite $3$ of them with condition that no couple won't be more then once on dinner. (When we took $A,B,C$ we cannot take $A$ with $B$ or $C$ and $B$ with $C$). My idea: We have $7$ subset with $3$ elements. So $7!$,but subset $\{A,B,C\}=\{B,A,C\}$ etc. So we have to divide by $3!$, but I don't think that's correct thinking. Should I divide also by couples? ($2!^3$) Any hints?

We can chose 3 men from 7 in $7$$C$$3$ many ways. That is $35$. Now, a couple repeats $5$ times. So, the we can chose three men with the criteria in $35/5=7$ many ways. Now, there are $7$ parties in total. So, we can permute the groups across days. That gives a total of $7!$ many ways. 
