Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am watching a lecture on pigeonhole principle at this link.

enter image description here

At time 40:42, why does the instructor say that "either a will have 3 friends or 3 enemies". Why can't it be any of the other cases she mentions ?


share|cite|improve this question

closed as off-topic by Andrés E. Caicedo, egreg, user127.0.0.1, studiosus, M Turgeon May 12 '14 at 21:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, egreg, user127.0.0.1, studiosus, M Turgeon
If this question can be reworded to fit the rules in the help center, please edit the question.

you can't pose questions like that. Explain what the problem is, then what you don't understand. You can't link to a video so someone has to watch it to understand the problem. – Ant Dec 17 '13 at 22:30
I did not watch all of it! 30 seconds at most :) – String Dec 17 '13 at 22:41
yeah i guess, but still.. The point of the site is that people can then search the same question and find the answer, in this way it's impossible ;-) – Ant Dec 17 '13 at 23:15

It can be, but in every case $a$ has at least $3$ friends or at least $3$ enemies, and that’s all that’s needed to make the rest of the argument work.

share|cite|improve this answer

She states the possible cases: $$ \begin{array}{c:c} \mbox{friends} & \mbox{enemies}\\ \hdashline\\ 5 & 0\\ \hdashline\\ 4 & 1\\ \hdashline\\ 3 & 2\\ \hdashline\\ 2 & 3\\ \hdashline\\ 1 & 4\\ \hdashline\\ 0 & 5 \end{array} $$ So what she should have said was that A would have at least three either friends or enemies. There is no row above having both numbers less than 3 is another way of putting it.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.