Combinatorial Proof for the equation $sum_{i=0}^j {j choose i} 2^{j-i} = 3^j$












1












$begingroup$


$$sum_{i=0}^j {j choose i}2^{j-i} = 3^j$$



My approach: I know the binomial way to do this is to think of the RHS as $(1+2)^j$ and then expand using binomial like so:
$$(1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i} cdot 1^i$$
$$ = (1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i}$$



But I am not sure how to do the combinatorial proof.






combinatorics discrete-mathematics binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by 'the combinatorial proof'?
    $endgroup$
    – caverac
    Dec 21 '18 at 10:36










  • $begingroup$
    @caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $kbinom{n}{k} = nbinom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people.
    $endgroup$
    – N. F. Taussig
    Dec 21 '18 at 10:46










  • $begingroup$
    @N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid
    $endgroup$
    – caverac
    Dec 21 '18 at 12:20
















1












$begingroup$


$$sum_{i=0}^j {j choose i}2^{j-i} = 3^j$$



My approach: I know the binomial way to do this is to think of the RHS as $(1+2)^j$ and then expand using binomial like so:
$$(1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i} cdot 1^i$$
$$ = (1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i}$$



But I am not sure how to do the combinatorial proof.






combinatorics discrete-mathematics binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by 'the combinatorial proof'?
    $endgroup$
    – caverac
    Dec 21 '18 at 10:36










  • $begingroup$
    @caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $kbinom{n}{k} = nbinom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people.
    $endgroup$
    – N. F. Taussig
    Dec 21 '18 at 10:46










  • $begingroup$
    @N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid
    $endgroup$
    – caverac
    Dec 21 '18 at 12:20














1












1








1


1



$begingroup$


$$sum_{i=0}^j {j choose i}2^{j-i} = 3^j$$



My approach: I know the binomial way to do this is to think of the RHS as $(1+2)^j$ and then expand using binomial like so:
$$(1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i} cdot 1^i$$
$$ = (1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i}$$



But I am not sure how to do the combinatorial proof.






combinatorics discrete-mathematics binomial-theorem







share|cite|improve this question











$endgroup$




$$sum_{i=0}^j {j choose i}2^{j-i} = 3^j$$



My approach: I know the binomial way to do this is to think of the RHS as $(1+2)^j$ and then expand using binomial like so:
$$(1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i} cdot 1^i$$
$$ = (1+2)^j = sum_{i=0}^j {j choose i} cdot 2^{j-i}$$



But I am not sure how to do the combinatorial proof.






combinatorics discrete-mathematics binomial-theorem




combinatorics discrete-mathematics binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 12:23









N. F. Taussig

45.2k103358




45.2k103358










asked Dec 21 '18 at 10:33









hussain sagarhussain sagar

908




908












  • $begingroup$
    What do you mean by 'the combinatorial proof'?
    $endgroup$
    – caverac
    Dec 21 '18 at 10:36










  • $begingroup$
    @caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $kbinom{n}{k} = nbinom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people.
    $endgroup$
    – N. F. Taussig
    Dec 21 '18 at 10:46










  • $begingroup$
    @N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid
    $endgroup$
    – caverac
    Dec 21 '18 at 12:20


















  • $begingroup$
    What do you mean by 'the combinatorial proof'?
    $endgroup$
    – caverac
    Dec 21 '18 at 10:36










  • $begingroup$
    @caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $kbinom{n}{k} = nbinom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people.
    $endgroup$
    – N. F. Taussig
    Dec 21 '18 at 10:46










  • $begingroup$
    @N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid
    $endgroup$
    – caverac
    Dec 21 '18 at 12:20
















$begingroup$
What do you mean by 'the combinatorial proof'?
$endgroup$
– caverac
Dec 21 '18 at 10:36




$begingroup$
What do you mean by 'the combinatorial proof'?
$endgroup$
– caverac
Dec 21 '18 at 10:36












$begingroup$
@caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $kbinom{n}{k} = nbinom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people.
$endgroup$
– N. F. Taussig
Dec 21 '18 at 10:46




$begingroup$
@caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $kbinom{n}{k} = nbinom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people.
$endgroup$
– N. F. Taussig
Dec 21 '18 at 10:46












$begingroup$
@N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid
$endgroup$
– caverac
Dec 21 '18 at 12:20




$begingroup$
@N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid
$endgroup$
– caverac
Dec 21 '18 at 12:20










2 Answers
2






active

oldest

votes


















7












$begingroup$

A combinatorical proof could go as follows:




  • The number of digit sequences of length $j$ formed with $3$ digits ${1,2,3}$ is: $color{blue}{3^j}$.

  • Now, fix one digit. For example $1$. It can occur $color{blue}{i=0,..,j}$ times in a digit sequence.

  • The number of ways to place $i$ times the digit $1$ is: $color{blue}{binom{j}{i}}$

  • You can fill the remaining $j-i$ places with any of the two other digits: $color{blue}{2^{j-i}}$
    All together:
    $$boxed{color{blue}{sum_{i=0}^j binom{j}{i}2^{j-i} = 3^j}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like your use of colour! :)
    $endgroup$
    – Shaun
    Dec 21 '18 at 11:27



















3












$begingroup$

The problem - how many $j$-length vectors can be composed of the digits ${0,1,2}$?



RHS - straight forward.



LHS - first, pick the $i$-indexes in the vector where 0 appears - $j choose i$, then choose between ${1,2}$ for the $j-i$ remaining indexes - $2^{j-i}$ options of doing so. Summing over all $i$'s gives all the required vectors.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    A combinatorical proof could go as follows:




    • The number of digit sequences of length $j$ formed with $3$ digits ${1,2,3}$ is: $color{blue}{3^j}$.

    • Now, fix one digit. For example $1$. It can occur $color{blue}{i=0,..,j}$ times in a digit sequence.

    • The number of ways to place $i$ times the digit $1$ is: $color{blue}{binom{j}{i}}$

    • You can fill the remaining $j-i$ places with any of the two other digits: $color{blue}{2^{j-i}}$
      All together:
      $$boxed{color{blue}{sum_{i=0}^j binom{j}{i}2^{j-i} = 3^j}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I like your use of colour! :)
      $endgroup$
      – Shaun
      Dec 21 '18 at 11:27
















    7












    $begingroup$

    A combinatorical proof could go as follows:




    • The number of digit sequences of length $j$ formed with $3$ digits ${1,2,3}$ is: $color{blue}{3^j}$.

    • Now, fix one digit. For example $1$. It can occur $color{blue}{i=0,..,j}$ times in a digit sequence.

    • The number of ways to place $i$ times the digit $1$ is: $color{blue}{binom{j}{i}}$

    • You can fill the remaining $j-i$ places with any of the two other digits: $color{blue}{2^{j-i}}$
      All together:
      $$boxed{color{blue}{sum_{i=0}^j binom{j}{i}2^{j-i} = 3^j}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I like your use of colour! :)
      $endgroup$
      – Shaun
      Dec 21 '18 at 11:27














    7












    7








    7





    $begingroup$

    A combinatorical proof could go as follows:




    • The number of digit sequences of length $j$ formed with $3$ digits ${1,2,3}$ is: $color{blue}{3^j}$.

    • Now, fix one digit. For example $1$. It can occur $color{blue}{i=0,..,j}$ times in a digit sequence.

    • The number of ways to place $i$ times the digit $1$ is: $color{blue}{binom{j}{i}}$

    • You can fill the remaining $j-i$ places with any of the two other digits: $color{blue}{2^{j-i}}$
      All together:
      $$boxed{color{blue}{sum_{i=0}^j binom{j}{i}2^{j-i} = 3^j}}$$






    share|cite|improve this answer









    $endgroup$



    A combinatorical proof could go as follows:




    • The number of digit sequences of length $j$ formed with $3$ digits ${1,2,3}$ is: $color{blue}{3^j}$.

    • Now, fix one digit. For example $1$. It can occur $color{blue}{i=0,..,j}$ times in a digit sequence.

    • The number of ways to place $i$ times the digit $1$ is: $color{blue}{binom{j}{i}}$

    • You can fill the remaining $j-i$ places with any of the two other digits: $color{blue}{2^{j-i}}$
      All together:
      $$boxed{color{blue}{sum_{i=0}^j binom{j}{i}2^{j-i} = 3^j}}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 10:50









    trancelocationtrancelocation

    14k1829




    14k1829












    • $begingroup$
      I like your use of colour! :)
      $endgroup$
      – Shaun
      Dec 21 '18 at 11:27


















    • $begingroup$
      I like your use of colour! :)
      $endgroup$
      – Shaun
      Dec 21 '18 at 11:27
















    $begingroup$
    I like your use of colour! :)
    $endgroup$
    – Shaun
    Dec 21 '18 at 11:27




    $begingroup$
    I like your use of colour! :)
    $endgroup$
    – Shaun
    Dec 21 '18 at 11:27











    3












    $begingroup$

    The problem - how many $j$-length vectors can be composed of the digits ${0,1,2}$?



    RHS - straight forward.



    LHS - first, pick the $i$-indexes in the vector where 0 appears - $j choose i$, then choose between ${1,2}$ for the $j-i$ remaining indexes - $2^{j-i}$ options of doing so. Summing over all $i$'s gives all the required vectors.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The problem - how many $j$-length vectors can be composed of the digits ${0,1,2}$?



      RHS - straight forward.



      LHS - first, pick the $i$-indexes in the vector where 0 appears - $j choose i$, then choose between ${1,2}$ for the $j-i$ remaining indexes - $2^{j-i}$ options of doing so. Summing over all $i$'s gives all the required vectors.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The problem - how many $j$-length vectors can be composed of the digits ${0,1,2}$?



        RHS - straight forward.



        LHS - first, pick the $i$-indexes in the vector where 0 appears - $j choose i$, then choose between ${1,2}$ for the $j-i$ remaining indexes - $2^{j-i}$ options of doing so. Summing over all $i$'s gives all the required vectors.






        share|cite|improve this answer









        $endgroup$



        The problem - how many $j$-length vectors can be composed of the digits ${0,1,2}$?



        RHS - straight forward.



        LHS - first, pick the $i$-indexes in the vector where 0 appears - $j choose i$, then choose between ${1,2}$ for the $j-i$ remaining indexes - $2^{j-i}$ options of doing so. Summing over all $i$'s gives all the required vectors.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 10:49









        BelkanBelkan

        687




        687






























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