At a meeting of nine mathematicians each pair converse in some language.No mathematician speak more than three languages. Show that there is some language spoken by at least five mathematician.
I don't know how to approach this kind of problem. Can anyone help?
Edit
I have posted another similar type of problem here .
Show that there is some language spoken by at least five
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combinatorics
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0Same language? How many are there in total? Right now this is trivially nonsensical. – 2017-01-31
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0I do not have an exactly idea, but the pigeonhole-principle should be helpful. – 2017-01-31
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2@ZirconCode I do not know a solution of the problem, but I think that we need not know how many languages there are. I think it will be enough to know that each pair of mathematimaticians have a common language and that no mathematician has more than $3$ languages. – 2017-01-31
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0Oh dear, yes you're right, apologies. – 2017-01-31
2 Answers
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Consider mathematician $A$, who speaks at most three languages and has to converse with eight other people. If he uses the same language with any four others, there are five who speak it. So the distribution is $3,3,2$. Now the languages he shares with $3$ others are already spoken by $4$. Can you use this to complete a proof?
Further hint - the distribution $3,3,2$ applies generally, so there can be no $1$s
Now this is not quite so easy as I thought when I put the hint, but this still makes good progress.
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Not sure if this is the direction, but at least some progress.
Denote by $m_{i}$ mathematicians and $l_{j}$ languages.
Take $m_{1}$. He needs to have a language to talk to $(m_{i})_{i=2}^{9}$. If $m_{1}$ speaks only one language, $l_{1}$, say, then all other mathematicians speak $l_{1}$, not possible. If $m_{1}$ speaks two languages, $l_{1}$ and $l_{2}$ say, then no matter how you distribute these two languages among $(m_{i})_{i=2}^{9}$, five mathematicians will speak either $l_{1}$ or $l_{2}$, not possible. Hence $m_{1}$ speaks three languages $l_{1}$, $l_{2}$ and $l_{3}$.
Since $m_{1}$ speaks $(l_{j})_{j=1}^{3}$, at most three $(m_{i})_{i=2}^{9}$ speak $l_{j}$ for all $j\in\{1,2,3\}$ (if four spoke $l_{j}$ then $m_{1}$ and these four would spoke $l_{j}$). Since there are eight other mathematicians besides $m_{1}$, one can assume without loss of generality that $(m_{i})_{i=2,3}$ speak $l_{3}$, $(m_{i})_{i=4,5,6}$ speak $l_{2}$ and $(m_{i})_{i=7,8,9}$ speak $l_{1}$.
Since $(m_{i})_{i=1,7,8,9}$ speak $l_{1}$, no other mathematician can speak $l_{1}$. Since $(m_{i})_{i=1,2,3,4}$ speak $l_{3}$, no other mathematician can speak $l_{3}$.