John has n acquaintances. He wants to meet 3 of them every day of 2018 for coffee at his home. What is the smallest value of n such that he can do this without calling the same set of three people more than once during 2018? Note that a person can come on multiple days, but never more than once with the same 2 other people. For the above n, we want to make a statement like “There is some acquaintance of John who came over at least m times”. What is the value of m given by the Pigeonhole principle?
So far this is what I have done to solve for n: $\frac{n!}{3!(n-3)!} = \frac{(n-1)(n-2)}{6}$, but I am unsure about how to find the smallest value of n for this problem. I am also unsure if I am calculating this correctly.
You are looking for a number $n$ such that there are at least 365 ways of choosing distinct groups of 3 people. Therefore, $$365\leq {n\choose 3}$$ And now search the smallest $n$ verifying that.
For $n=14$ the rhs. evaluates to 364, so either you let John take a day off, or take $n=15$ with 455 different tea meetings.
Regarding your other question, the value of $m$ counting how many meetings did a particular person attended to can be analyzed as follows. Say that Mary went to $m$ of the 365 meetings. This separes the meetings into two disjoint sets, the first $m$ meetings in which Mary was present, and the resting $365-m$.
For the first $m$ meetings, we need to choose 2 people out of $n-1$ to organize those meetings, so we have $$m\leq{n-1\choose 2}.$$ For the rest of the meetings, the argument is as before: $$365-m\leq{n-1\choose 3}$$
Therefore, $m$ must verify the following:
$$365-{n-1\choose 3}\leq m \leq {n-1\choose 2}$$
For $n=15$ this is $1\leq m\leq 91$. So everyone has to attend at least once, and no more than 91 times.