Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$...












2












$begingroup$


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?






combinatorics pigeonhole-principle







share|cite|improve this question











$endgroup$












  • $begingroup$
    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 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 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 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.
    $endgroup$
    – Supakorn Srisawat
    Jan 27 at 19:16








  • 1




    $begingroup$
    If you can't fit your question into 150 characters, you should explain it in the body of the question.
    $endgroup$
    – TonyK
    Jan 27 at 19:55
















2












$begingroup$


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?






combinatorics pigeonhole-principle







share|cite|improve this question











$endgroup$












  • $begingroup$
    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 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 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 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.
    $endgroup$
    – Supakorn Srisawat
    Jan 27 at 19:16








  • 1




    $begingroup$
    If you can't fit your question into 150 characters, you should explain it in the body of the question.
    $endgroup$
    – TonyK
    Jan 27 at 19:55














2












2








2


1



$begingroup$


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?






combinatorics pigeonhole-principle







share|cite|improve this question











$endgroup$




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?






combinatorics pigeonhole-principle




combinatorics pigeonhole-principle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 19:07







Supakorn Srisawat

















asked Jan 27 at 18:56









Supakorn SrisawatSupakorn Srisawat

625




625












  • $begingroup$
    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 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 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 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.
    $endgroup$
    – Supakorn Srisawat
    Jan 27 at 19:16








  • 1




    $begingroup$
    If you can't fit your question into 150 characters, you should explain it in the body of the question.
    $endgroup$
    – TonyK
    Jan 27 at 19:55


















  • $begingroup$
    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 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 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 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.
    $endgroup$
    – Supakorn Srisawat
    Jan 27 at 19:16








  • 1




    $begingroup$
    If you can't fit your question into 150 characters, you should explain it in the body of the question.
    $endgroup$
    – TonyK
    Jan 27 at 19:55
















$begingroup$
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 at 19:07






$begingroup$
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 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 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 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 at 19:10




$begingroup$
I think it is trying to say that there are 5 distinct inhabited rows, or 5 distinct inhabited columns.
$endgroup$
– DanielV
Jan 27 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.
$endgroup$
– Supakorn Srisawat
Jan 27 at 19:16






$begingroup$
@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.
$endgroup$
– Supakorn Srisawat
Jan 27 at 19:16






1




1




$begingroup$
If you can't fit your question into 150 characters, you should explain it in the body of the question.
$endgroup$
– TonyK
Jan 27 at 19:55




$begingroup$
If you can't fit your question into 150 characters, you should explain it in the body of the question.
$endgroup$
– TonyK
Jan 27 at 19:55










2 Answers
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$begingroup$

$4$ 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)






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    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






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      $begingroup$

      $4$ 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)






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        $4$ 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)






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          $4$ 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)






          share|cite|improve this answer











          $endgroup$



          $4$ 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)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 27 at 19:33

























          answered Jan 27 at 19:07









          Bram28Bram28

          63.9k44793




          63.9k44793























              2












              $begingroup$

              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






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                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






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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






                  share|cite|improve this answer









                  $endgroup$



                  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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 27 at 19:56









                  Maria MazurMaria Mazur

                  48.3k1260121




                  48.3k1260121






























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