Does there exist a similar identity to $binom{n}{k} = binom{n}{n-k}$?












1














I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients







share|cite|improve this question




















  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12


















1














I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients







share|cite|improve this question




















  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12
















1












1








1







I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients







share|cite|improve this question















I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients




combinatorics soft-question binomial-coefficients






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 3:41









Shaun

8,507113580




8,507113580










asked Nov 24 at 2:57









Shrey Joshi

19512




19512








  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12
















  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12










2




2




No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
– David Peterson
Nov 24 at 3:12






No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
– David Peterson
Nov 24 at 3:12












2 Answers
2






active

oldest

votes


















2














It is standard to define
$$
binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
$$
as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






share|cite|improve this answer





























    0














    No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



    An algebraic proof of the same fact is the following-



    Multiply both LHS and RHS by $k!$, we get,
    $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
    Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



    We can see this directly from the Pascal's identity-
    $$binom nk+binom n{k-1} =binom {n+1}k$$
    $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



    Hope this is helpful.






    share|cite|improve this answer

















    • 1




      What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
      – Shrey Joshi
      Nov 24 at 3:37











    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%2f3011121%2fdoes-there-exist-a-similar-identity-to-binomnk-binomnn-k%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














    It is standard to define
    $$
    binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
    $$
    as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






    share|cite|improve this answer


























      2














      It is standard to define
      $$
      binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
      $$
      as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






      share|cite|improve this answer
























        2












        2








        2






        It is standard to define
        $$
        binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
        $$
        as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






        share|cite|improve this answer












        It is standard to define
        $$
        binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
        $$
        as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 4:09









        Micah

        29.6k1363105




        29.6k1363105























            0














            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.






            share|cite|improve this answer

















            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37
















            0














            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.






            share|cite|improve this answer

















            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37














            0












            0








            0






            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.






            share|cite|improve this answer












            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 3:13









            Martund

            1,351212




            1,351212








            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37














            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37








            1




            1




            What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
            – Shrey Joshi
            Nov 24 at 3:37




            What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
            – Shrey Joshi
            Nov 24 at 3:37


















            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3011121%2fdoes-there-exist-a-similar-identity-to-binomnk-binomnn-k%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more