Part a, 208/30=7. See?.
I am stuck on this question involving the pigeon hole principle. Any help is appreciated.
In a gathering of 30 people, there are 104 different pairs of people who know each other.
a. Show that some person must have at least seven acquaintances.
b. Show that some person must have fewer than seven acquaintances.
I think this one uses the ceiling function, but I am confused.
Actually the pigeons are the acquaintances and there are 208.
Each pair contributes two acquaintances. The holes are the persons.
If each hole had at most six pigeons see the contradiction?
If each hole had at least seven pigeons see the contradiction?
BTW: using graph theory makes this much easier.