# Pigeon hole principle application [closed]

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

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 ?

Thanks.

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

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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.
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.