Combinatoric proof that $sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$












3












$begingroup$


Proof that:




$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.




Can someone help me understand this equality by giving a combinatoric argument?



I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.



I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.



Thanks for helping!






combinatorics







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  • $begingroup$
    What have you tried?
    $endgroup$
    – Phicar
    Jan 22 at 2:52
















3












$begingroup$


Proof that:




$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.




Can someone help me understand this equality by giving a combinatoric argument?



I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.



I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.



Thanks for helping!






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – Phicar
    Jan 22 at 2:52














3












3








3


0



$begingroup$


Proof that:




$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.




Can someone help me understand this equality by giving a combinatoric argument?



I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.



I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.



Thanks for helping!






combinatorics







share|cite|improve this question











$endgroup$




Proof that:




$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.




Can someone help me understand this equality by giving a combinatoric argument?



I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.



I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.



Thanks for helping!






combinatorics




combinatorics






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share|cite|improve this question













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edited Jan 22 at 19:09









Blue

48.7k870156




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asked Jan 20 at 20:28









ZacharyZachary

1799




1799












  • $begingroup$
    What have you tried?
    $endgroup$
    – Phicar
    Jan 22 at 2:52


















  • $begingroup$
    What have you tried?
    $endgroup$
    – Phicar
    Jan 22 at 2:52
















$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52




$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52










1 Answer
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Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.






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    $begingroup$

    Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.






        share|cite|improve this answer









        $endgroup$



        Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 19:02









        PhicarPhicar

        2,6401915




        2,6401915






























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