Combinatorial proof for a sum of binomial coefficient products [duplicate]

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  • Prove that $sumlimits_k=0^mbinommkbinomn+km=sumlimits_k=0^mbinomnkbinommk2^k$ [duplicate]

    4 answers



  • Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$

    2 answers



I would like to prove the following statement to be true.
$$
sum_k = 0^m binommkbinomn+km
= sum_k = 0^mbinomnkbinommk2^k
$$







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marked as duplicate by amWhy, Jendrik Stelzner, N. F. Taussig combinatorics
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Aug 26 at 11:16


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  • First, I would try to express the combinatorial expressions as fractions.
    – DavidS
    Aug 26 at 11:08














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  • Prove that $sumlimits_k=0^mbinommkbinomn+km=sumlimits_k=0^mbinomnkbinommk2^k$ [duplicate]

    4 answers



  • Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$

    2 answers



I would like to prove the following statement to be true.
$$
sum_k = 0^m binommkbinomn+km
= sum_k = 0^mbinomnkbinommk2^k
$$







share|cite|improve this question














marked as duplicate by amWhy, Jendrik Stelzner, N. F. Taussig combinatorics
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  • First, I would try to express the combinatorial expressions as fractions.
    – DavidS
    Aug 26 at 11:08












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0
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up vote
0
down vote

favorite












This question already has an answer here:



  • Prove that $sumlimits_k=0^mbinommkbinomn+km=sumlimits_k=0^mbinomnkbinommk2^k$ [duplicate]

    4 answers



  • Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$

    2 answers



I would like to prove the following statement to be true.
$$
sum_k = 0^m binommkbinomn+km
= sum_k = 0^mbinomnkbinommk2^k
$$







share|cite|improve this question















This question already has an answer here:



  • Prove that $sumlimits_k=0^mbinommkbinomn+km=sumlimits_k=0^mbinomnkbinommk2^k$ [duplicate]

    4 answers



  • Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$

    2 answers



I would like to prove the following statement to be true.
$$
sum_k = 0^m binommkbinomn+km
= sum_k = 0^mbinomnkbinommk2^k
$$





This question already has an answer here:



  • Prove that $sumlimits_k=0^mbinommkbinomn+km=sumlimits_k=0^mbinomnkbinommk2^k$ [duplicate]

    4 answers



  • Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$

    2 answers









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 26 at 11:08









Jendrik Stelzner

7,58221037




7,58221037










asked Aug 26 at 11:05









Alan DSouza

6




6




marked as duplicate by amWhy, Jendrik Stelzner, N. F. Taussig combinatorics
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marked as duplicate by amWhy, Jendrik Stelzner, N. F. Taussig combinatorics
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • First, I would try to express the combinatorial expressions as fractions.
    – DavidS
    Aug 26 at 11:08
















  • First, I would try to express the combinatorial expressions as fractions.
    – DavidS
    Aug 26 at 11:08















First, I would try to express the combinatorial expressions as fractions.
– DavidS
Aug 26 at 11:08




First, I would try to express the combinatorial expressions as fractions.
– DavidS
Aug 26 at 11:08















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