Probability that sum of three digits is the same as sum of other three digits












1














How can I solve problems like this:



Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,



$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.



I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.






probability combinatorics







share|cite|improve this question
























  • Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
    – Keen-ameteur
    Nov 18 at 14:55






  • 1




    Are the numbers chosen with or without replacement?
    – Ross Millikan
    Nov 18 at 14:57










  • Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 18 at 15:25
















1














How can I solve problems like this:



Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,



$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.



I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.






probability combinatorics







share|cite|improve this question
























  • Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
    – Keen-ameteur
    Nov 18 at 14:55






  • 1




    Are the numbers chosen with or without replacement?
    – Ross Millikan
    Nov 18 at 14:57










  • Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 18 at 15:25














1












1








1







How can I solve problems like this:



Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,



$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.



I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.






probability combinatorics







share|cite|improve this question















How can I solve problems like this:



Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,



$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.



I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.






probability combinatorics




probability combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 at 16:11









N. F. Taussig

43.5k93355




43.5k93355










asked Nov 18 at 14:51









Gokuruto

123




123












  • Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
    – Keen-ameteur
    Nov 18 at 14:55






  • 1




    Are the numbers chosen with or without replacement?
    – Ross Millikan
    Nov 18 at 14:57










  • Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 18 at 15:25


















  • Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
    – Keen-ameteur
    Nov 18 at 14:55






  • 1




    Are the numbers chosen with or without replacement?
    – Ross Millikan
    Nov 18 at 14:57










  • Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 18 at 15:25
















Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55




Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55




1




1




Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57




Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57












Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25




Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25










2 Answers
2






active

oldest

votes


















2














It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$






share|cite|improve this answer





























    0














    You are looking for
    $$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
    0 leqslant text{integer }x_{,j} leqslant r hfill \
    x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
    end{gathered} right.$$



    which is given by
    $$
    N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
    sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
    {left( { - 1} right)^k binom{m}{k}
    binom
    { s + m - 1 - kleft( {r + 1} right) }
    { s - kleft( {r + 1} right)} }
    $$
    as widely explained in this related post.



    Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.






    share|cite|improve this answer





















      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%2f3003625%2fprobability-that-sum-of-three-digits-is-the-same-as-sum-of-other-three-digits%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 seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
      The final result $p$ is then simply given by
      $$p=sum_{k=0}^{27}p_k^2 .$$
      By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
      $$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$






      share|cite|improve this answer


























        2














        It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
        The final result $p$ is then simply given by
        $$p=sum_{k=0}^{27}p_k^2 .$$
        By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
        $$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$






        share|cite|improve this answer
























          2












          2








          2






          It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
          The final result $p$ is then simply given by
          $$p=sum_{k=0}^{27}p_k^2 .$$
          By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
          $$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$






          share|cite|improve this answer












          It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
          The final result $p$ is then simply given by
          $$p=sum_{k=0}^{27}p_k^2 .$$
          By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
          $$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 16:49









          Christian Blatter

          172k7112325




          172k7112325























              0














              You are looking for
              $$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
              0 leqslant text{integer }x_{,j} leqslant r hfill \
              x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
              end{gathered} right.$$



              which is given by
              $$
              N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
              sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
              {left( { - 1} right)^k binom{m}{k}
              binom
              { s + m - 1 - kleft( {r + 1} right) }
              { s - kleft( {r + 1} right)} }
              $$
              as widely explained in this related post.



              Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.






              share|cite|improve this answer


























                0














                You are looking for
                $$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
                0 leqslant text{integer }x_{,j} leqslant r hfill \
                x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
                end{gathered} right.$$



                which is given by
                $$
                N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
                sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
                {left( { - 1} right)^k binom{m}{k}
                binom
                { s + m - 1 - kleft( {r + 1} right) }
                { s - kleft( {r + 1} right)} }
                $$
                as widely explained in this related post.



                Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  You are looking for
                  $$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
                  0 leqslant text{integer }x_{,j} leqslant r hfill \
                  x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
                  end{gathered} right.$$



                  which is given by
                  $$
                  N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
                  sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
                  {left( { - 1} right)^k binom{m}{k}
                  binom
                  { s + m - 1 - kleft( {r + 1} right) }
                  { s - kleft( {r + 1} right)} }
                  $$
                  as widely explained in this related post.



                  Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.






                  share|cite|improve this answer












                  You are looking for
                  $$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
                  0 leqslant text{integer }x_{,j} leqslant r hfill \
                  x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
                  end{gathered} right.$$



                  which is given by
                  $$
                  N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
                  sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
                  {left( { - 1} right)^k binom{m}{k}
                  binom
                  { s + m - 1 - kleft( {r + 1} right) }
                  { s - kleft( {r + 1} right)} }
                  $$
                  as widely explained in this related post.



                  Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 18 at 17:08









                  G Cab

                  17.9k31237




                  17.9k31237






























                      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%2f3003625%2fprobability-that-sum-of-three-digits-is-the-same-as-sum-of-other-three-digits%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

                      AnyDesk - Fatal Program Failure

                      Image
                      up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
                      Read more

                      How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                      Image
                      up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
                      Read more

                      QoS: MAC-Priority for clients behind a repeater

                      Image
                      1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
                      Read more