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. 
 A: We will solve this using the Inclusion-Exclusion Principle.
Since the person has $7$ friends, each night he/she has $\binom{7}{3}$ possible groups of friends he/she can invite over. Notice he/she will do this for $7$ nights, but the groups have to be different every night. Thus we have to consider the permutations of $7$ groups of friends (from the possible $\binom{7}{3}$ groups). This yields $P\left(\binom{7}{3},7\right) $. This is the total number of ways for the person to invite different groups of $3$ friends each night, for $7$ nights.
Now we have to subtract the number of cases where the person omits one friend. Using exactly the same arguments as before, there is a total of $P\left(\binom{6}{3},7\right)$ ways to do so. Since we can choose the friend to omit in $\binom{7}{1} = 7$ ways, we have to subtract $ \ \ 7 \times P\left(\binom{6}{3},7\right)$ to the total.
Now we add the number of ways where the person omits two friends. By the same reasoning as before, this turns out to be $\binom{7}{2}\times  P\left(\binom{5}{3},7\right)$. 
Notice that the next step is to subtract the case where the person omits $3$ of their friends. But then we would be left to choose from $4$ friends to invite. There are only $\binom{4}{3} = 4$ different groups of $3$ we can form, and we need to have at least $7$ (one for every night), so we can stop here.
So, by the Inclusion-Exclusion Principle, the answer is given by
$$P\left(\binom{7}{3},7\right) - 7 \cdot P\left(\binom{6}{3},7\right) + \binom{7}{2}\cdot  P\left(\binom{5}{3},7\right).$$ 
A: 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.....
$$\frac{35!}{28!}-7\cdot \frac{20!}{13!} + 21\cdot \frac{10!}{3!}$$
