Person with seven friends invites subset of three for one week Suppose that a person with seven friends invites a different subset of three friends to dinner every night for one week (seven days). How many ways can this be done so that all friends are included at least once?
I know the number of ways to do this with no restriction is $\binom{7}{3}^7$. 
I need to use Inclusion-Exclusion..however I am having trouble determining the sets to use. 

This answer is essentially the same as Thomas Bladt's but I use a different format for counting. $$\frac {\binom{7}{3}!}{(\binom{7}{3}-7)!} - 7\cdot \frac{(\binom{7}{3}-\binom{6}{2})!}{(\binom{7}{3}-\binom{6}{2}-7)!}+ \binom{7}{2}\cdot \frac{(\binom{7}{3}-\binom{6}{2}-\binom{5}{2})!}{(\binom{7}{3}-\binom{6}{2}-\binom{5}{2}-7)!}$$ Which reduces to.....
$$\frac{35!}{28!}-7\cdot \frac{20!}{13!} + 21\cdot \frac{10!}{3!}$$
