A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
How do I prove the following:
In a room of 100 people, show that at least 3 people have the same number of acquaintances, where the number of acquaintances has to be even (including 0).
anonymous
 5 years ago
How do I prove the following: In a room of 100 people, show that at least 3 people have the same number of acquaintances, where the number of acquaintances has to be even (including 0).

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I came up with a solution: Uses pigeonhole principle  Case 1: Only one person has 0 friends, therefore taking that person out of the number of possible friends. Thus the remaining 99 people must have between 2, 4, 6, ... , 98 friends, i.e. 49 possibilities for for the 99 people. Since 2*49 + 1 = 99, there are at least 3 people with the same number of friends. Case 2: Only 2 people have 0 friends (if I said three, the proof becomes trivial). The remaining 98 peopel have between 2, 4, 6, .... , 96 friends, i.e. 48 possibilities. Since 2*48 + 2 = 98, there are at least 3 people with the same number of friends. Case 3: Everyone has at least 1 friend. But since everyone have to have an even number of friends, they have between 2 , 4 , 6, ... , 98 friends, i.e. 49 possibilities per person. Since 2*49+2 = 100, at least 3 people have the same number of friends. Make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I came up with a solution: Uses pigeonhole principle  Case 1: Only one person has 0 friends, therefore taking that person out of the number of possible friends. Thus the remaining 99 people must have between 2, 4, 6, ... , 98 friends, i.e. 49 possibilities for for the 99 people. Since 2*49 + 1 = 99, there are at least 3 people with the same number of friends. Case 2: Only 2 people have 0 friends (if I said three, the proof becomes trivial). The remaining 98 peopel have between 2, 4, 6, .... , 96 friends, i.e. 48 possibilities. Since 2*48 + 2 = 98, there are at least 3 people with the same number of friends. Case 3: Everyone has at least 1 friend. But since everyone have to have an even number of friends, they have between 2 , 4 , 6, ... , 98 friends, i.e. 49 possibilities per person. Since 2*49+2 = 100, at least 3 people have the same number of friends. Make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I came up with a solution: Uses pigeonhole principle  Case 1: Only one person has 0 friends, therefore taking that person out of the number of possible friends. Thus the remaining 99 people must have between 2, 4, 6, ... , 98 friends, i.e. 49 possibilities for for the 99 people. Since 2*49 + 1 = 99, there are at least 3 people with the same number of friends. Case 2: Only 2 people have 0 friends (if I said three, the proof becomes trivial). The remaining 98 peopel have between 2, 4, 6, .... , 96 friends, i.e. 48 possibilities. Since 2*48 + 2 = 98, there are at least 3 people with the same number of friends. Case 3: Everyone has at least 1 friend. But since everyone have to have an even number of friends, they have between 2 , 4 , 6, ... , 98 friends, i.e. 49 possibilities per person. Since 2*49+2 = 100, at least 3 people have the same number of friends. Make sense?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.