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In a more quantified version: for natural numbers $k$ and $m$, if $n = km + 1$ objects are distributed among $m$ sets, then the pigeonhole principle asserts that atleast one of the sets will contain at least $k + 1$ objects. (from wikipedia) Now why is 'atleast' there? Only one set should contain only $k+1$ objects, as $n$ is just one more than $k$ items (are being contained by $m$ sets) , i.e. $n=km+1$

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  • $\begingroup$ Not following. Maybe one set contains all the objects and all the others are empty. Or, if that example is missing your point, can you give a numerical example of what is bothering you? $\endgroup$
    – lulu
    May 12, 2018 at 18:19
  • $\begingroup$ No I am not able to grasp the actual essence of that line. Can you please give me example. On wikipedia there is example of 10 pigeons 9 holes so that is satisfied but any examples in favor of atleast $\endgroup$ May 12, 2018 at 18:23
  • $\begingroup$ An example of what? I don't understand. If I have $2$ objects and $1$ hole then exactly $1$ of the holes has more than $1$ object. Is that an example of what you want? $\endgroup$
    – lulu
    May 12, 2018 at 18:32
  • $\begingroup$ Thank you Sir for your help, I have got my answer what I wanted. $\endgroup$ May 12, 2018 at 18:45

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Let $k=3$ and $m=2$ so that $n=7$. Then one possible arrangement is $(5,2)$, but neither of these is $k+1=4$.

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  • $\begingroup$ Thank you, sir. (Oh that was quite easy, just because of little confusion I asked that silly question, sorry I bothered you). But, Thanks for your help,Sir. $\endgroup$ May 12, 2018 at 18:40

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