retagged by
2,001 views

2 Answers

Best answer
4 votes
4 votes
0 votes
0 votes

In a group of N people, a person may have 0, 1, 2, ..., N − 1 friends. Assume to the contrary that all N people have different number of friends. Then for each number in the sequence 0, 1, 2, ..., N − 1 there must be a person with exactly this number of friends. In particular, there is at least one with N − 1 friends. But, if so, all others have this person as a friend, implying that there is no one with no friends at all. Therefore, the only possible numbers of friends come from the shortened sequence: 1, 2, 3, ..., N − 1. By the Pigeonhole Principle, there are at least two with the same number of friends, so that our assumption that this is not true proved wrong, thus it must be indeed true.

https://gateoverflow.in/338531/kenneth-rosen-edition-7th-exercise-6-question-40-page-no-406

Related questions

0 votes
0 votes
1 answer
1
aditi19 asked Oct 25, 2018
981 views
How many cards must be chosen from a standard deck of 52 cards to guarantee that there are at least two cards of each of two different kinds?what this question means?
1 votes
1 votes
1 answer
2
Sammohan Ganguly asked May 25, 2018
1,364 views
Suppose a graph $G$ has $6$ nodes. Prove that either $G$ or $G'$ must contain a triangle.($G'$ is the complement of $G$.) Prove it using pigeonhole principle.
4 votes
4 votes
2 answers
3
sourav. asked Aug 24, 2016
2,179 views
Show that if seven integers are selected from the first10 positive integers, there must be at least two pairsof these integers with the sum 11.Attempt-:partition will be ...
0 votes
0 votes
1 answer
4
Anu asked Jul 14, 2015
2,052 views
Show that there are at least six people in California (population: 37 million) with the same three initials who were born on the same day of the year (but not necessarily...