Question

At a party of n people, some have a symmetric friendship. Symmetric means that if a is friends with b, then b is in turn friends with
a. Prove that there are at-least two people with same number of friends.

Answer 1

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As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).

As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Solution #1:

# of selections of 2 out of 7 - ;

# of selections of 2 people which are not siblings - one from first pair of siblings*one from second pair of siblings + one from first pair of siblings*one from triple + one from second pair of siblings*one from triple .

= 4 + 6 + 6

= 16

P = 16/21

Answer 2

Proof:

The number of friends can range from 0 to n-1. If someone has 0 friends, then no one can have n-1 friends, and similarly, if someone has n-1 friends, then no one can have 0 friends. So real possibilities of the number of friends of a person can be n-1 (excluding the case of 0 or n-1). And there are total n people. So, according to the pigeon hole principal, any two people must have the same number of friends.