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?
 A: Either I am missing something or the book answer is wrong. I take it that when you write "no couple won't be more than once on dinner", you actually mean "no couple will ....." as has been amply clarified in your examples.
Now there can only be $\dbinom72 = 21$ distinct couples, and each day $3$ such couples get eliminated, e.g. if $ABC$ are invited, $AB, BC$ and $AC$ can't be invited again.
Thus there are only $7$ trios to be invited, and if the order in which they are invited is to be considered, there are $7!$ ways of inviting them.
Note
There is no question of dividing by $3!$, because we considered combos of $3$, and for the life of me, I can't understand multiplication by $2$  and division by $2^2$] 
A: 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. 
