Prove the identity $sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} = 3^n$












1












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Prove the identity $sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} = 3^n$. I believe I need to use the binomial theorem here, but I don't know how to deal with the double summations.






combinatorics discrete-mathematics summation binomial-coefficients







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  • $begingroup$
    Use this question.
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    – Dietrich Burde
    Dec 8 '18 at 19:19










  • $begingroup$
    You may be interested in the Multinomial theorem about multinomial coefficients.
    $endgroup$
    – Somos
    Dec 8 '18 at 19:33
















1












$begingroup$


Prove the identity $sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} = 3^n$. I believe I need to use the binomial theorem here, but I don't know how to deal with the double summations.






combinatorics discrete-mathematics summation binomial-coefficients







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use this question.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:19










  • $begingroup$
    You may be interested in the Multinomial theorem about multinomial coefficients.
    $endgroup$
    – Somos
    Dec 8 '18 at 19:33














1












1








1


0



$begingroup$


Prove the identity $sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} = 3^n$. I believe I need to use the binomial theorem here, but I don't know how to deal with the double summations.






combinatorics discrete-mathematics summation binomial-coefficients







share|cite|improve this question











$endgroup$




Prove the identity $sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} = 3^n$. I believe I need to use the binomial theorem here, but I don't know how to deal with the double summations.






combinatorics discrete-mathematics summation binomial-coefficients




combinatorics discrete-mathematics summation binomial-coefficients






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 9 '18 at 15:48









Martin Sleziak

44.7k9117272




44.7k9117272










asked Dec 8 '18 at 19:13









cosmicbrowniecosmicbrownie

1016




1016












  • $begingroup$
    Use this question.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:19










  • $begingroup$
    You may be interested in the Multinomial theorem about multinomial coefficients.
    $endgroup$
    – Somos
    Dec 8 '18 at 19:33


















  • $begingroup$
    Use this question.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:19










  • $begingroup$
    You may be interested in the Multinomial theorem about multinomial coefficients.
    $endgroup$
    – Somos
    Dec 8 '18 at 19:33
















$begingroup$
Use this question.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:19




$begingroup$
Use this question.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:19












$begingroup$
You may be interested in the Multinomial theorem about multinomial coefficients.
$endgroup$
– Somos
Dec 8 '18 at 19:33




$begingroup$
You may be interested in the Multinomial theorem about multinomial coefficients.
$endgroup$
– Somos
Dec 8 '18 at 19:33










4 Answers
4






active

oldest

votes


















2












$begingroup$

We will use the identity
$$
binom{n}{k}binom{k}{r}=binom{n}{r}binom{n-r}{k-r}quad (ngeq kgeq rgeq 0).
$$
Interchange the order of summation and use the identity above to get that
$$
sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k}
=sum_{r=0}^{n}binom{n}{r}sum_{k=r}^{n}binom{n-r}{k-r}=sum_{r=0}^nbinom{n}{r}2^{n-r}=(1+2)^n=3^n
$$
where we used the fact that
$$
sum_{k=r}^{n}binom{n-r}{k-r}=sum_{u=0}^{n-r}binom{n-r}{u}=2^{n-r}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you know a combinatorial proof of the statement?
    $endgroup$
    – nafhgood
    Dec 8 '18 at 20:33






  • 1




    $begingroup$
    @mathnoob I think I have a combinatorial argument, see my answer below.
    $endgroup$
    – Ned
    Dec 8 '18 at 21:18










  • $begingroup$
    That's cool! Thanks!
    $endgroup$
    – nafhgood
    Dec 8 '18 at 21:22



















2












$begingroup$

Let $|X|=n$, so $3^n$ is the number of functions from $X$ to {$1,2,3$}.



Each such function corresponds uniquely to a pair of subsets $(A,B)$ with $A$ a subset of $B$ and $B$ a subset of $X$ by taking $B$ = {$x$ | $f(x)=2$ or $f(x)=3$} and $A$ = {$x$ | $f(x)=2$}.



The number of such pairs of nested subsets of $X$ is the double sum on the right hand side of the formula (where $k$ = $|B|$ and $r$ = $|A|$).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You know that $displaystyle(1+x)^n=sum_{k=0}^nbinom nkx^k$



    $displaystylesum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} =sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}=sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}1^r$



    Since $displaystylesum_{r=0}^{k} binom{k}{r}1^r=(1+1)^k=2^k$, we have



    $displaystyleimpliessum_{k=0}^{n} binom nk2^k=(1+2)^n=3^n$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$


      A convenient aspect is the index of the inner sum affects only one binomial coefficient. Setting parenthesis we might observe



      begin{align*}
      color{blue}{sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r}binom{n}{k}}
      &=sum_{k=0}^{n}left(sum_{r=0}^{k} binom{k}{r} right)binom{n}{k}\
      &=sum_{k=0}^{n}2^k binom{n}{k}\
      &,,color{blue}{=3^n}
      end{align*}







      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        We will use the identity
        $$
        binom{n}{k}binom{k}{r}=binom{n}{r}binom{n-r}{k-r}quad (ngeq kgeq rgeq 0).
        $$
        Interchange the order of summation and use the identity above to get that
        $$
        sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k}
        =sum_{r=0}^{n}binom{n}{r}sum_{k=r}^{n}binom{n-r}{k-r}=sum_{r=0}^nbinom{n}{r}2^{n-r}=(1+2)^n=3^n
        $$
        where we used the fact that
        $$
        sum_{k=r}^{n}binom{n-r}{k-r}=sum_{u=0}^{n-r}binom{n-r}{u}=2^{n-r}.
        $$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Do you know a combinatorial proof of the statement?
          $endgroup$
          – nafhgood
          Dec 8 '18 at 20:33






        • 1




          $begingroup$
          @mathnoob I think I have a combinatorial argument, see my answer below.
          $endgroup$
          – Ned
          Dec 8 '18 at 21:18










        • $begingroup$
          That's cool! Thanks!
          $endgroup$
          – nafhgood
          Dec 8 '18 at 21:22
















        2












        $begingroup$

        We will use the identity
        $$
        binom{n}{k}binom{k}{r}=binom{n}{r}binom{n-r}{k-r}quad (ngeq kgeq rgeq 0).
        $$
        Interchange the order of summation and use the identity above to get that
        $$
        sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k}
        =sum_{r=0}^{n}binom{n}{r}sum_{k=r}^{n}binom{n-r}{k-r}=sum_{r=0}^nbinom{n}{r}2^{n-r}=(1+2)^n=3^n
        $$
        where we used the fact that
        $$
        sum_{k=r}^{n}binom{n-r}{k-r}=sum_{u=0}^{n-r}binom{n-r}{u}=2^{n-r}.
        $$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Do you know a combinatorial proof of the statement?
          $endgroup$
          – nafhgood
          Dec 8 '18 at 20:33






        • 1




          $begingroup$
          @mathnoob I think I have a combinatorial argument, see my answer below.
          $endgroup$
          – Ned
          Dec 8 '18 at 21:18










        • $begingroup$
          That's cool! Thanks!
          $endgroup$
          – nafhgood
          Dec 8 '18 at 21:22














        2












        2








        2





        $begingroup$

        We will use the identity
        $$
        binom{n}{k}binom{k}{r}=binom{n}{r}binom{n-r}{k-r}quad (ngeq kgeq rgeq 0).
        $$
        Interchange the order of summation and use the identity above to get that
        $$
        sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k}
        =sum_{r=0}^{n}binom{n}{r}sum_{k=r}^{n}binom{n-r}{k-r}=sum_{r=0}^nbinom{n}{r}2^{n-r}=(1+2)^n=3^n
        $$
        where we used the fact that
        $$
        sum_{k=r}^{n}binom{n-r}{k-r}=sum_{u=0}^{n-r}binom{n-r}{u}=2^{n-r}.
        $$






        share|cite|improve this answer









        $endgroup$



        We will use the identity
        $$
        binom{n}{k}binom{k}{r}=binom{n}{r}binom{n-r}{k-r}quad (ngeq kgeq rgeq 0).
        $$
        Interchange the order of summation and use the identity above to get that
        $$
        sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k}
        =sum_{r=0}^{n}binom{n}{r}sum_{k=r}^{n}binom{n-r}{k-r}=sum_{r=0}^nbinom{n}{r}2^{n-r}=(1+2)^n=3^n
        $$
        where we used the fact that
        $$
        sum_{k=r}^{n}binom{n-r}{k-r}=sum_{u=0}^{n-r}binom{n-r}{u}=2^{n-r}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 19:22









        Foobaz JohnFoobaz John

        21.8k41352




        21.8k41352












        • $begingroup$
          Do you know a combinatorial proof of the statement?
          $endgroup$
          – nafhgood
          Dec 8 '18 at 20:33






        • 1




          $begingroup$
          @mathnoob I think I have a combinatorial argument, see my answer below.
          $endgroup$
          – Ned
          Dec 8 '18 at 21:18










        • $begingroup$
          That's cool! Thanks!
          $endgroup$
          – nafhgood
          Dec 8 '18 at 21:22


















        • $begingroup$
          Do you know a combinatorial proof of the statement?
          $endgroup$
          – nafhgood
          Dec 8 '18 at 20:33






        • 1




          $begingroup$
          @mathnoob I think I have a combinatorial argument, see my answer below.
          $endgroup$
          – Ned
          Dec 8 '18 at 21:18










        • $begingroup$
          That's cool! Thanks!
          $endgroup$
          – nafhgood
          Dec 8 '18 at 21:22
















        $begingroup$
        Do you know a combinatorial proof of the statement?
        $endgroup$
        – nafhgood
        Dec 8 '18 at 20:33




        $begingroup$
        Do you know a combinatorial proof of the statement?
        $endgroup$
        – nafhgood
        Dec 8 '18 at 20:33




        1




        1




        $begingroup$
        @mathnoob I think I have a combinatorial argument, see my answer below.
        $endgroup$
        – Ned
        Dec 8 '18 at 21:18




        $begingroup$
        @mathnoob I think I have a combinatorial argument, see my answer below.
        $endgroup$
        – Ned
        Dec 8 '18 at 21:18












        $begingroup$
        That's cool! Thanks!
        $endgroup$
        – nafhgood
        Dec 8 '18 at 21:22




        $begingroup$
        That's cool! Thanks!
        $endgroup$
        – nafhgood
        Dec 8 '18 at 21:22











        2












        $begingroup$

        Let $|X|=n$, so $3^n$ is the number of functions from $X$ to {$1,2,3$}.



        Each such function corresponds uniquely to a pair of subsets $(A,B)$ with $A$ a subset of $B$ and $B$ a subset of $X$ by taking $B$ = {$x$ | $f(x)=2$ or $f(x)=3$} and $A$ = {$x$ | $f(x)=2$}.



        The number of such pairs of nested subsets of $X$ is the double sum on the right hand side of the formula (where $k$ = $|B|$ and $r$ = $|A|$).






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Let $|X|=n$, so $3^n$ is the number of functions from $X$ to {$1,2,3$}.



          Each such function corresponds uniquely to a pair of subsets $(A,B)$ with $A$ a subset of $B$ and $B$ a subset of $X$ by taking $B$ = {$x$ | $f(x)=2$ or $f(x)=3$} and $A$ = {$x$ | $f(x)=2$}.



          The number of such pairs of nested subsets of $X$ is the double sum on the right hand side of the formula (where $k$ = $|B|$ and $r$ = $|A|$).






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Let $|X|=n$, so $3^n$ is the number of functions from $X$ to {$1,2,3$}.



            Each such function corresponds uniquely to a pair of subsets $(A,B)$ with $A$ a subset of $B$ and $B$ a subset of $X$ by taking $B$ = {$x$ | $f(x)=2$ or $f(x)=3$} and $A$ = {$x$ | $f(x)=2$}.



            The number of such pairs of nested subsets of $X$ is the double sum on the right hand side of the formula (where $k$ = $|B|$ and $r$ = $|A|$).






            share|cite|improve this answer









            $endgroup$



            Let $|X|=n$, so $3^n$ is the number of functions from $X$ to {$1,2,3$}.



            Each such function corresponds uniquely to a pair of subsets $(A,B)$ with $A$ a subset of $B$ and $B$ a subset of $X$ by taking $B$ = {$x$ | $f(x)=2$ or $f(x)=3$} and $A$ = {$x$ | $f(x)=2$}.



            The number of such pairs of nested subsets of $X$ is the double sum on the right hand side of the formula (where $k$ = $|B|$ and $r$ = $|A|$).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '18 at 21:11









            NedNed

            1,993910




            1,993910























                1












                $begingroup$

                You know that $displaystyle(1+x)^n=sum_{k=0}^nbinom nkx^k$



                $displaystylesum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} =sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}=sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}1^r$



                Since $displaystylesum_{r=0}^{k} binom{k}{r}1^r=(1+1)^k=2^k$, we have



                $displaystyleimpliessum_{k=0}^{n} binom nk2^k=(1+2)^n=3^n$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  You know that $displaystyle(1+x)^n=sum_{k=0}^nbinom nkx^k$



                  $displaystylesum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} =sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}=sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}1^r$



                  Since $displaystylesum_{r=0}^{k} binom{k}{r}1^r=(1+1)^k=2^k$, we have



                  $displaystyleimpliessum_{k=0}^{n} binom nk2^k=(1+2)^n=3^n$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    You know that $displaystyle(1+x)^n=sum_{k=0}^nbinom nkx^k$



                    $displaystylesum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} =sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}=sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}1^r$



                    Since $displaystylesum_{r=0}^{k} binom{k}{r}1^r=(1+1)^k=2^k$, we have



                    $displaystyleimpliessum_{k=0}^{n} binom nk2^k=(1+2)^n=3^n$






                    share|cite|improve this answer









                    $endgroup$



                    You know that $displaystyle(1+x)^n=sum_{k=0}^nbinom nkx^k$



                    $displaystylesum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r} binom{n}{k} =sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}=sum_{k=0}^{n}binom{n}{k}sum_{r=0}^{k} binom{k}{r}1^r$



                    Since $displaystylesum_{r=0}^{k} binom{k}{r}1^r=(1+1)^k=2^k$, we have



                    $displaystyleimpliessum_{k=0}^{n} binom nk2^k=(1+2)^n=3^n$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 8 '18 at 19:31









                    Shubham JohriShubham Johri

                    4,992717




                    4,992717























                        0












                        $begingroup$


                        A convenient aspect is the index of the inner sum affects only one binomial coefficient. Setting parenthesis we might observe



                        begin{align*}
                        color{blue}{sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r}binom{n}{k}}
                        &=sum_{k=0}^{n}left(sum_{r=0}^{k} binom{k}{r} right)binom{n}{k}\
                        &=sum_{k=0}^{n}2^k binom{n}{k}\
                        &,,color{blue}{=3^n}
                        end{align*}







                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$


                          A convenient aspect is the index of the inner sum affects only one binomial coefficient. Setting parenthesis we might observe



                          begin{align*}
                          color{blue}{sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r}binom{n}{k}}
                          &=sum_{k=0}^{n}left(sum_{r=0}^{k} binom{k}{r} right)binom{n}{k}\
                          &=sum_{k=0}^{n}2^k binom{n}{k}\
                          &,,color{blue}{=3^n}
                          end{align*}







                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$


                            A convenient aspect is the index of the inner sum affects only one binomial coefficient. Setting parenthesis we might observe



                            begin{align*}
                            color{blue}{sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r}binom{n}{k}}
                            &=sum_{k=0}^{n}left(sum_{r=0}^{k} binom{k}{r} right)binom{n}{k}\
                            &=sum_{k=0}^{n}2^k binom{n}{k}\
                            &,,color{blue}{=3^n}
                            end{align*}







                            share|cite|improve this answer









                            $endgroup$




                            A convenient aspect is the index of the inner sum affects only one binomial coefficient. Setting parenthesis we might observe



                            begin{align*}
                            color{blue}{sum_{k=0}^{n}sum_{r=0}^{k} binom{k}{r}binom{n}{k}}
                            &=sum_{k=0}^{n}left(sum_{r=0}^{k} binom{k}{r} right)binom{n}{k}\
                            &=sum_{k=0}^{n}2^k binom{n}{k}\
                            &,,color{blue}{=3^n}
                            end{align*}








                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 9 '18 at 14:54









                            Markus ScheuerMarkus Scheuer

                            60.8k456145




                            60.8k456145






























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                                Tonle Sap (See)

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                                Tonle Sap Satellitenfoto (April 1994) Geographische Lage Kambodscha Zuflüsse Tonle-Sap-Fluss, Siem Reap River Abfluss Tonle-Sap-Fluss Orte in der Nähe Siem Reap, Battambang Daten Koordinaten 12° 52′ 34″  N , 104° 4′ 23″  O 12.876 104.073 Koordinaten: 12° 52′ 34″  N , 104° 4′ 23″  O Höhe über Meeresspiegel f1 0,5 Meter Fläche 2.700–25.000 km² dep1 f5 Länge 250 km (maximal) dep1 f6 Breite 100 km (maximal) dep1 f7 Volumen 80 km³ (maximal) dep1 f8 Maximale Tiefe 14 m f10 Mittlere Tiefe 2–3 m (minimal) f11 Besonderheiten größter See Südostasiens Der Tonle-Sap-See (Khmer: បឹង​ទន្លេសាប , Aussprache: [ boeung tunleː saːp ]) in Kambodscha ist der größte See Südostasiens und eines der fischreichsten Binnengewässer der Erde. Inhaltsverzeichnis 1 Geographie 2 Wirtschaftliche Bedeutung 2.1 Verkehr 2.2 Reis 2.3 Fisch 3 Ökologie und Gefährdung 3.1 Überfischung...
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                                I get strange results when I access the Sqlitedatabase with Unity C# via XAMPP

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                                0 <?php $servername = "localhost"; $username="root"; $password=""; $dbName="escape_room"; $user_username = $_POST['Input_user']; $user_password = $_POST['Input_pass']; $conn = new mysqli($servername, $username, $password, $dbName); if(!$conn){ die("Cound not Connect: " . mysqli_connect_error()); } $sql = "SELECT pass FROM escape_room WHERE user = '".$user_username."' "; $result = mysqli_query($conn, $sql); //Get the result and confirm login if(mysqli_num_rows($result)>0){ while($row = mysqli_fetch_assoc($result)){ if($row['pass'] == $user_password){ echo "login success"; ...
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                                Guatemaltekische Davis-Cup-Mannschaft

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                                Guatemala Kapitän Anthony Vásquez Aktuelles ITF-Ranking 73 Statistik Erste Teilnahme 1990 Davis-Cup-Teilnahmen 22 Bestes Ergebnis Amerika-Gruppenzone II HF (1995, 2000, 2001) Ewige Bilanz 42:30 Erfolgreichste Spieler Meiste Siege gesamt Daniel Chávez Morales (29) Meiste Einzelsiege Cristian Paiz (21) Meiste Doppelsiege Daniel Chávez Morales (17) Bestes Doppel Jacobo Chávez / Daniel Chávez Morales (9) Meiste Teilnahmen Cristian Paiz (32) Meiste Jahre Daniel Chávez Morales (10) Letzte Aktualisierung der Infobox: 3. Juni 2012 Die guatemaltekische Davis-Cup-Mannschaft ist die Tennisnationalmannschaft Guatemalas. Geschichte | 1990 nahm Guatemala erstmals am Davis Cup teil. Das beste Resultat erreichte die Mannschaft jeweils 1995, 2000 und 2001 mit dem Halbfinale in der Amerika-Gruppenzone II. Bester Spieler ist Daniel Chávez Morales mit 29 Siegen bei insgesamt 31 Teilnahmen. Rekord...
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