Question

There are 101 people at a party. Each person has an even number (possibly zero) of acquaintances. Prove that there are three people at the party with the same number of acquaintances. Is this true if there are only 100 people? Justify your answer. Note that a person cannot be an acquaintance with themself.

Answer 1

Hint  •There are 100 people at a party.
•Each has even number (possible 0) of friends
•Prove that we can always find three people
with the same number of friends  Proof: There are three cases.
Case 1: At most 1 person has 0 friends
# friends for each remaining person is from 2 to 98
By Generalized Pigeonhole Principle,
at least three persons have same # friends
Case 2: Exactly two persons has 0 friends
# friends for each remaining person is from 2 to 96
Case 3: At least three persons has 0 friends (Trivially) Therom use -(The Pigeonhole Principle). If k ∈ Z

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+ and k + 1 or more objects are placed into k boxes, then

there is at least one box containing two or more of the objects.

Proof. We prove the pigeonhole principle using a proof by contraposition. Suppose that none of the k boxes

contains more than one object. Then the total number of objects would be at most k. This is a contradiction,

because there are at least k + 1 objects.