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There are two ways to prove it. One way is...

Consider $41$ chess pieces on $10$ board rows. By using pigeonhole princ., there must be one row that has at least $\lceil{\frac{41}{10}}\rceil$ $=$ $5$ (but less than $10$) pieces. Eliminate that row. Then there are $9$ rows and at least $31$ chess pieces remaining. By using pigeonhole princ., there must be one row that has at least $\lceil{\frac{31}{9}}\rceil$ $=$ $4$ pieces..........

As you can see, this proving way is really long, you have to eliminate rows by rows and choose one piece in one of the eliminated rows to prove and conclude.

Can you find the other way to prove by using the pigeonhole principle for only one time?

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  • $\begingroup$ not following. There must be at least five non-empty rows, just choose one from each of those. Am i missing something? $\endgroup$
    – lulu
    Jan 27 '19 at 19:07
  • $\begingroup$ I can't make sense of your title: "there are at least $5$...that are not on the same row." What does that even mean? $\endgroup$
    – TonyK
    Jan 27 '19 at 19:07
  • $\begingroup$ I think it is trying to say that there are 5 distinct inhabited rows, or 5 distinct inhabited columns. $\endgroup$
    – DanielV
    Jan 27 '19 at 19:10
  • $\begingroup$ @DanielV You're right. The question in my textbook was actually said (in Thai), at least $5$ pieces that are not in the same row or column. But the $150$ letters rule make it ambiguous for you. $\endgroup$ Jan 27 '19 at 19:16
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    $\begingroup$ If you can't fit your question into 150 characters, you should explain it in the body of the question. $\endgroup$
    – TonyK
    Jan 27 '19 at 19:55
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$4$ rows can at most provide room for $4 \cdot 10 = 40$ pieces, So, with $41$ pieces, you'll need need to use at least $5$ rows to fit all the pieces, meaning that there have to be $5$ pieces that are on $5$ different rows (I assume that that is what the question is asking you to prove)

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Since we have $10$ colulmns we must have, by Pigeonhole principle, in one of them at least $5$ chess pices. So they are not in the same row and we are done

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