Thread: Pigeon Holes Question

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.

Part a, 208/30=7. See?.

Originally Posted by galactus

Part a, 208/30=7. See?.

I feel really stupid now. Thank you. I must be tired.

For part b, maybe try a contradiction argument of some sort. Often, that is a good approach when doing a PHP proof.

Maybe show there is a contradiction by there being at least 105 pairs of acquaintances. That may be one way to approach it.

Originally Posted by galactus

For part b, maybe try a contradiction argument of some sort. Often, that is a good approach when doing a PHP proof.

Maybe show there is a contradiction by there being at least 105 pairs of acquaintances. That may be one way to approach it.

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.