Suppose Dianne has a number of friends. She wants to invite three to her house for dinner every day of the year. What is the minimum number of friends she can have if she does not want to invite the same three friends more than once?
I reason that the number of friends can be obtained by $ {n \choose 3} = \frac{n!}{3!(n-3)!} = 365$
but this gives me a third-order polynomial which is unsolvable without a computer. I know guess and check works but I am looking for a "more correct" solution that uses mathematics. I know this si also technically incorrect because you cant "choose" out of numbers other then integers. I know the solution but I'm having trouble framing my answer using combinatorics
I also tried applying pigeonhole principle to this question (initially) but that didn't get me anywhere.
It is interesting to notice that you face the same problem as Newton before Newton method !
Simpliying all numbers, you need to solve for $n$ the cubic equatio $$n^3-3 n^2+2 n-2190=0$$
First guess : $n^3-2190=0 \implies n=\sqrt[3]{2190}=12.9862$; say $n=13$.
Now, make one single iteration of Newton method to get $n=\frac{6077}{431}=14.0998$. So $n=\lceil 14.0998\rceil=15$.