Help finding a combinatorial proof of ${kn choose 2}= k{n choose 2}+n^2{k choose 2}$ [closed]












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Hi I have been trying to find a way to find a combinatorial proof for ${kn choose 2}= k{n choose 2}+n^2{k choose 2}$.






combinatorics discrete-mathematics







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closed as off-topic by RRL, jgon, KReiser, Cesareo, amWhy Jan 5 at 1:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, jgon, KReiser, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.


















    5












    $begingroup$


    Hi I have been trying to find a way to find a combinatorial proof for ${kn choose 2}= k{n choose 2}+n^2{k choose 2}$.






    combinatorics discrete-mathematics







    share|cite|improve this question









    $endgroup$



    closed as off-topic by RRL, jgon, KReiser, Cesareo, amWhy Jan 5 at 1:57


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, jgon, KReiser, Cesareo, amWhy

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      5












      5








      5


      3



      $begingroup$


      Hi I have been trying to find a way to find a combinatorial proof for ${kn choose 2}= k{n choose 2}+n^2{k choose 2}$.






      combinatorics discrete-mathematics







      share|cite|improve this question









      $endgroup$




      Hi I have been trying to find a way to find a combinatorial proof for ${kn choose 2}= k{n choose 2}+n^2{k choose 2}$.






      combinatorics discrete-mathematics




      combinatorics discrete-mathematics






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      share|cite|improve this question











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      asked Dec 6 '18 at 23:28









      Lauren SmithLauren Smith

      291




      291




      closed as off-topic by RRL, jgon, KReiser, Cesareo, amWhy Jan 5 at 1:57


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, jgon, KReiser, Cesareo, amWhy

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by RRL, jgon, KReiser, Cesareo, amWhy Jan 5 at 1:57


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, jgon, KReiser, Cesareo, amWhy

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$

          Hint: You want to pick $2$ elements out of $k$ buckets of $n$ elements each. You have two possible ways to do it: either you pick a bucket, and then you take $2$ elements from this bucket, or you pick $2$ buckets and then you choose one element from each of those $2$ buckets.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:38



















          3












          $begingroup$

          Consider a set of $kcdot n$ elements allocated in a grid with $n$ rows and $k$ columns then




          • on the LHS we have the ways to choose $2$ elements among all of them

          • on the RHS we have the cases with 2 elements choseen from a same row $k{n choose 2}$ and the cases 2 elements choseen from a same column $n{k choose 2}$ and the orther cases with $2$ elements chosen form different columns and rows $nk+(n-1)(k-1)$, indeed


          $$k{n choose 2}+n{k choose 2}+nk+(n-1)(k-1)=k{n choose 2}+n^2{k choose 2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:39



















          1












          $begingroup$

          As an addendum to Daniel Robert-Nicoud’s excellent hint, I find it helpful to rewrite the equality in the following way:



          $$binom{kn}{2} = binom{k}{1}binom{n}{2} + binom{k}{2}binom{n}{1}binom{n}{1}.$$



          By reading multiplication as “and then,” and addition as “or,” try to recover Daniel’s narrative by reading off the right-hand side.




          With this thought process in mind, try to come up with a formula for $binom{nk}{3}$. Can you generalize this pattern?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:37


















          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Hint: You want to pick $2$ elements out of $k$ buckets of $n$ elements each. You have two possible ways to do it: either you pick a bucket, and then you take $2$ elements from this bucket, or you pick $2$ buckets and then you choose one element from each of those $2$ buckets.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:38
















          6












          $begingroup$

          Hint: You want to pick $2$ elements out of $k$ buckets of $n$ elements each. You have two possible ways to do it: either you pick a bucket, and then you take $2$ elements from this bucket, or you pick $2$ buckets and then you choose one element from each of those $2$ buckets.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:38














          6












          6








          6





          $begingroup$

          Hint: You want to pick $2$ elements out of $k$ buckets of $n$ elements each. You have two possible ways to do it: either you pick a bucket, and then you take $2$ elements from this bucket, or you pick $2$ buckets and then you choose one element from each of those $2$ buckets.






          share|cite|improve this answer









          $endgroup$



          Hint: You want to pick $2$ elements out of $k$ buckets of $n$ elements each. You have two possible ways to do it: either you pick a bucket, and then you take $2$ elements from this bucket, or you pick $2$ buckets and then you choose one element from each of those $2$ buckets.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 23:30









          Daniel Robert-NicoudDaniel Robert-Nicoud

          20.3k33596




          20.3k33596












          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:38


















          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:38
















          $begingroup$
          Thank you very much. This was very helpful.
          $endgroup$
          – Lauren Smith
          Dec 7 '18 at 7:38




          $begingroup$
          Thank you very much. This was very helpful.
          $endgroup$
          – Lauren Smith
          Dec 7 '18 at 7:38











          3












          $begingroup$

          Consider a set of $kcdot n$ elements allocated in a grid with $n$ rows and $k$ columns then




          • on the LHS we have the ways to choose $2$ elements among all of them

          • on the RHS we have the cases with 2 elements choseen from a same row $k{n choose 2}$ and the cases 2 elements choseen from a same column $n{k choose 2}$ and the orther cases with $2$ elements chosen form different columns and rows $nk+(n-1)(k-1)$, indeed


          $$k{n choose 2}+n{k choose 2}+nk+(n-1)(k-1)=k{n choose 2}+n^2{k choose 2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:39
















          3












          $begingroup$

          Consider a set of $kcdot n$ elements allocated in a grid with $n$ rows and $k$ columns then




          • on the LHS we have the ways to choose $2$ elements among all of them

          • on the RHS we have the cases with 2 elements choseen from a same row $k{n choose 2}$ and the cases 2 elements choseen from a same column $n{k choose 2}$ and the orther cases with $2$ elements chosen form different columns and rows $nk+(n-1)(k-1)$, indeed


          $$k{n choose 2}+n{k choose 2}+nk+(n-1)(k-1)=k{n choose 2}+n^2{k choose 2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:39














          3












          3








          3





          $begingroup$

          Consider a set of $kcdot n$ elements allocated in a grid with $n$ rows and $k$ columns then




          • on the LHS we have the ways to choose $2$ elements among all of them

          • on the RHS we have the cases with 2 elements choseen from a same row $k{n choose 2}$ and the cases 2 elements choseen from a same column $n{k choose 2}$ and the orther cases with $2$ elements chosen form different columns and rows $nk+(n-1)(k-1)$, indeed


          $$k{n choose 2}+n{k choose 2}+nk+(n-1)(k-1)=k{n choose 2}+n^2{k choose 2}$$






          share|cite|improve this answer









          $endgroup$



          Consider a set of $kcdot n$ elements allocated in a grid with $n$ rows and $k$ columns then




          • on the LHS we have the ways to choose $2$ elements among all of them

          • on the RHS we have the cases with 2 elements choseen from a same row $k{n choose 2}$ and the cases 2 elements choseen from a same column $n{k choose 2}$ and the orther cases with $2$ elements chosen form different columns and rows $nk+(n-1)(k-1)$, indeed


          $$k{n choose 2}+n{k choose 2}+nk+(n-1)(k-1)=k{n choose 2}+n^2{k choose 2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 23:46









          gimusigimusi

          1




          1












          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:39


















          • $begingroup$
            Thank you very much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:39
















          $begingroup$
          Thank you very much. This was very helpful.
          $endgroup$
          – Lauren Smith
          Dec 7 '18 at 7:39




          $begingroup$
          Thank you very much. This was very helpful.
          $endgroup$
          – Lauren Smith
          Dec 7 '18 at 7:39











          1












          $begingroup$

          As an addendum to Daniel Robert-Nicoud’s excellent hint, I find it helpful to rewrite the equality in the following way:



          $$binom{kn}{2} = binom{k}{1}binom{n}{2} + binom{k}{2}binom{n}{1}binom{n}{1}.$$



          By reading multiplication as “and then,” and addition as “or,” try to recover Daniel’s narrative by reading off the right-hand side.




          With this thought process in mind, try to come up with a formula for $binom{nk}{3}$. Can you generalize this pattern?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:37
















          1












          $begingroup$

          As an addendum to Daniel Robert-Nicoud’s excellent hint, I find it helpful to rewrite the equality in the following way:



          $$binom{kn}{2} = binom{k}{1}binom{n}{2} + binom{k}{2}binom{n}{1}binom{n}{1}.$$



          By reading multiplication as “and then,” and addition as “or,” try to recover Daniel’s narrative by reading off the right-hand side.




          With this thought process in mind, try to come up with a formula for $binom{nk}{3}$. Can you generalize this pattern?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:37














          1












          1








          1





          $begingroup$

          As an addendum to Daniel Robert-Nicoud’s excellent hint, I find it helpful to rewrite the equality in the following way:



          $$binom{kn}{2} = binom{k}{1}binom{n}{2} + binom{k}{2}binom{n}{1}binom{n}{1}.$$



          By reading multiplication as “and then,” and addition as “or,” try to recover Daniel’s narrative by reading off the right-hand side.




          With this thought process in mind, try to come up with a formula for $binom{nk}{3}$. Can you generalize this pattern?






          share|cite|improve this answer











          $endgroup$



          As an addendum to Daniel Robert-Nicoud’s excellent hint, I find it helpful to rewrite the equality in the following way:



          $$binom{kn}{2} = binom{k}{1}binom{n}{2} + binom{k}{2}binom{n}{1}binom{n}{1}.$$



          By reading multiplication as “and then,” and addition as “or,” try to recover Daniel’s narrative by reading off the right-hand side.




          With this thought process in mind, try to come up with a formula for $binom{nk}{3}$. Can you generalize this pattern?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 0:21

























          answered Dec 7 '18 at 0:05









          Santana AftonSantana Afton

          2,5752629




          2,5752629












          • $begingroup$
            Thank you so much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:37


















          • $begingroup$
            Thank you so much. This was very helpful.
            $endgroup$
            – Lauren Smith
            Dec 7 '18 at 7:37
















          $begingroup$
          Thank you so much. This was very helpful.
          $endgroup$
          – Lauren Smith
          Dec 7 '18 at 7:37




          $begingroup$
          Thank you so much. This was very helpful.
          $endgroup$
          – Lauren Smith
          Dec 7 '18 at 7:37



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