Prove that $sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} {^iC_j}=3^n$












1












$begingroup$


Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53
















1












$begingroup$


Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53














1












1








1


3



$begingroup$


Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations







share|cite|improve this question









$endgroup$




Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations




combinatorics discrete-mathematics combinations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 15:47









user1942348user1942348

1,3901934




1,3901934








  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53














  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53








1




1




$begingroup$
Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
$endgroup$
– darij grinberg
Jan 4 at 15:53




$begingroup$
Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
$endgroup$
– darij grinberg
Jan 4 at 15:53










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hints:




  • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

  • $3=1+2$

  • $2=1+1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
    $endgroup$
    – user1942348
    Jan 4 at 16:53












  • $begingroup$
    @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
    $endgroup$
    – Mike Earnest
    Jan 4 at 18:57










  • $begingroup$
    @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
    $endgroup$
    – Arthur
    Jan 4 at 18:59



















1












$begingroup$

First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




  1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

  2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


(So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



$$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






share|cite|improve this answer











$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061776%2fprove-that-sum-j-0n-sum-i-jn-nc-i-ic-j-3n%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59
















    2












    $begingroup$

    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59














    2












    2








    2





    $begingroup$

    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$






    share|cite|improve this answer









    $endgroup$



    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 15:58









    ArthurArthur

    122k7122211




    122k7122211












    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59


















    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59
















    $begingroup$
    Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
    $endgroup$
    – user1942348
    Jan 4 at 16:53






    $begingroup$
    Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
    $endgroup$
    – user1942348
    Jan 4 at 16:53














    $begingroup$
    @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
    $endgroup$
    – Mike Earnest
    Jan 4 at 18:57




    $begingroup$
    @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
    $endgroup$
    – Mike Earnest
    Jan 4 at 18:57












    $begingroup$
    @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
    $endgroup$
    – Arthur
    Jan 4 at 18:59




    $begingroup$
    @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
    $endgroup$
    – Arthur
    Jan 4 at 18:59











    1












    $begingroup$

    First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




    1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

    2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


    (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



    On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



    $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




      1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

      2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


      (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



      On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



      $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




        1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

        2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


        (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



        On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



        $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






        share|cite|improve this answer











        $endgroup$



        First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




        1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

        2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


        (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



        On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



        $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 19:52

























        answered Jan 4 at 19:31









        MikeMike

        4,611512




        4,611512






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061776%2fprove-that-sum-j-0n-sum-i-jn-nc-i-ic-j-3n%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more