How many $4$-digit lock combinations are possible if each digit in the code may appear at most twice?












1












$begingroup$


Question
A lock combination code is made up of $4$ numbers ($0-9$). Each number can occur at most twice, e.g. $4764$ would be allowed but not $4464$ as the number $4$ has occurred more than $2$ times.Therefore, how many possible combination codes are there?



I know that there are $10,000$ possible combinations if repetitions are allowed. However I'm unsure as to how to answer the question. Any help is greatly appreciated, thanks!






combinatorics discrete-mathematics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does "Each number can occur two times mostly." mean? Do you mean a number can occur $0,1$ or $2$ times?
    $endgroup$
    – lulu
    Dec 14 '18 at 0:27










  • $begingroup$
    @lulu For example '4764' would be allowed but not '4464' as the number '4' has occurred more than 2 times.
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 0:31










  • $begingroup$
    Got it. I think, then, you meant to say "Each number can occur at most two times."
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    As a hint: break it into three cases. Case I: no number repeats. Case 2: exactly one number repeats. Case 3: two numbers repeat (as in $3443$).
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    The complement method may be quicker than finding it straight up: find the number of ways you have $3$ digits the same and $4$ digits the same and subtract from $10000$
    $endgroup$
    – WaveX
    Dec 14 '18 at 0:33
















1












$begingroup$


Question
A lock combination code is made up of $4$ numbers ($0-9$). Each number can occur at most twice, e.g. $4764$ would be allowed but not $4464$ as the number $4$ has occurred more than $2$ times.Therefore, how many possible combination codes are there?



I know that there are $10,000$ possible combinations if repetitions are allowed. However I'm unsure as to how to answer the question. Any help is greatly appreciated, thanks!






combinatorics discrete-mathematics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does "Each number can occur two times mostly." mean? Do you mean a number can occur $0,1$ or $2$ times?
    $endgroup$
    – lulu
    Dec 14 '18 at 0:27










  • $begingroup$
    @lulu For example '4764' would be allowed but not '4464' as the number '4' has occurred more than 2 times.
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 0:31










  • $begingroup$
    Got it. I think, then, you meant to say "Each number can occur at most two times."
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    As a hint: break it into three cases. Case I: no number repeats. Case 2: exactly one number repeats. Case 3: two numbers repeat (as in $3443$).
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    The complement method may be quicker than finding it straight up: find the number of ways you have $3$ digits the same and $4$ digits the same and subtract from $10000$
    $endgroup$
    – WaveX
    Dec 14 '18 at 0:33














1












1








1


1



$begingroup$


Question
A lock combination code is made up of $4$ numbers ($0-9$). Each number can occur at most twice, e.g. $4764$ would be allowed but not $4464$ as the number $4$ has occurred more than $2$ times.Therefore, how many possible combination codes are there?



I know that there are $10,000$ possible combinations if repetitions are allowed. However I'm unsure as to how to answer the question. Any help is greatly appreciated, thanks!






combinatorics discrete-mathematics permutations







share|cite|improve this question











$endgroup$




Question
A lock combination code is made up of $4$ numbers ($0-9$). Each number can occur at most twice, e.g. $4764$ would be allowed but not $4464$ as the number $4$ has occurred more than $2$ times.Therefore, how many possible combination codes are there?



I know that there are $10,000$ possible combinations if repetitions are allowed. However I'm unsure as to how to answer the question. Any help is greatly appreciated, thanks!






combinatorics discrete-mathematics permutations




combinatorics discrete-mathematics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 9:22









N. F. Taussig

43.7k93355




43.7k93355










asked Dec 14 '18 at 0:26









Jasmine078Jasmine078

123




123












  • $begingroup$
    What does "Each number can occur two times mostly." mean? Do you mean a number can occur $0,1$ or $2$ times?
    $endgroup$
    – lulu
    Dec 14 '18 at 0:27










  • $begingroup$
    @lulu For example '4764' would be allowed but not '4464' as the number '4' has occurred more than 2 times.
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 0:31










  • $begingroup$
    Got it. I think, then, you meant to say "Each number can occur at most two times."
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    As a hint: break it into three cases. Case I: no number repeats. Case 2: exactly one number repeats. Case 3: two numbers repeat (as in $3443$).
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    The complement method may be quicker than finding it straight up: find the number of ways you have $3$ digits the same and $4$ digits the same and subtract from $10000$
    $endgroup$
    – WaveX
    Dec 14 '18 at 0:33


















  • $begingroup$
    What does "Each number can occur two times mostly." mean? Do you mean a number can occur $0,1$ or $2$ times?
    $endgroup$
    – lulu
    Dec 14 '18 at 0:27










  • $begingroup$
    @lulu For example '4764' would be allowed but not '4464' as the number '4' has occurred more than 2 times.
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 0:31










  • $begingroup$
    Got it. I think, then, you meant to say "Each number can occur at most two times."
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    As a hint: break it into three cases. Case I: no number repeats. Case 2: exactly one number repeats. Case 3: two numbers repeat (as in $3443$).
    $endgroup$
    – lulu
    Dec 14 '18 at 0:32






  • 2




    $begingroup$
    The complement method may be quicker than finding it straight up: find the number of ways you have $3$ digits the same and $4$ digits the same and subtract from $10000$
    $endgroup$
    – WaveX
    Dec 14 '18 at 0:33
















$begingroup$
What does "Each number can occur two times mostly." mean? Do you mean a number can occur $0,1$ or $2$ times?
$endgroup$
– lulu
Dec 14 '18 at 0:27




$begingroup$
What does "Each number can occur two times mostly." mean? Do you mean a number can occur $0,1$ or $2$ times?
$endgroup$
– lulu
Dec 14 '18 at 0:27












$begingroup$
@lulu For example '4764' would be allowed but not '4464' as the number '4' has occurred more than 2 times.
$endgroup$
– Jasmine078
Dec 14 '18 at 0:31




$begingroup$
@lulu For example '4764' would be allowed but not '4464' as the number '4' has occurred more than 2 times.
$endgroup$
– Jasmine078
Dec 14 '18 at 0:31












$begingroup$
Got it. I think, then, you meant to say "Each number can occur at most two times."
$endgroup$
– lulu
Dec 14 '18 at 0:32




$begingroup$
Got it. I think, then, you meant to say "Each number can occur at most two times."
$endgroup$
– lulu
Dec 14 '18 at 0:32




2




2




$begingroup$
As a hint: break it into three cases. Case I: no number repeats. Case 2: exactly one number repeats. Case 3: two numbers repeat (as in $3443$).
$endgroup$
– lulu
Dec 14 '18 at 0:32




$begingroup$
As a hint: break it into three cases. Case I: no number repeats. Case 2: exactly one number repeats. Case 3: two numbers repeat (as in $3443$).
$endgroup$
– lulu
Dec 14 '18 at 0:32




2




2




$begingroup$
The complement method may be quicker than finding it straight up: find the number of ways you have $3$ digits the same and $4$ digits the same and subtract from $10000$
$endgroup$
– WaveX
Dec 14 '18 at 0:33




$begingroup$
The complement method may be quicker than finding it straight up: find the number of ways you have $3$ digits the same and $4$ digits the same and subtract from $10000$
$endgroup$
– WaveX
Dec 14 '18 at 0:33










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here is a possible solution using the Principle of Inclusion/Exclusion:



$textbf{Case 1}$ (digit is repeated $3$ times):



Choose the repeated digit in $10 choose 1$ ways.



Choose the remaining digit in $9 choose 1$ ways.



Order the digits in $frac{4!}{3!} = 4$ ways.



$textbf{Case 2}$ (digit is repeated $4$ times):



Choose the repeated digit in $10 choose 1$ ways.



Order the digits in $frac{4!}{4!} = 1$ way.



So the number of combinations that don't satisfy your constraint is $(10 * 9 * 4) + (10 * 1) = 370$. Subtracting these from the $10,000$ total combinations yields $textbf{9,630}$ ways.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Solution without using the Principle of Inclusion/Exclusion:



    Add by number of pairs of numbers: $0$ pairs $+$ $1$ pair $+$ $2$ pairs



    $=$ (pick $4$ of the $10$ order matters) $+$
    (pick which will be repeated and where and the other $2$ numbers order matters) $+$
    (pick number for each pair and location of first and second pair)



    $=P(10,4) + 10*{{4}choose{2}}*P(9,2) + {{10}choose{2}}*{{4}choose{2}}*1$



    $=5040$ $+$ $4320$ $+$ $290$



    $= 9630$






    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%2f3038773%2fhow-many-4-digit-lock-combinations-are-possible-if-each-digit-in-the-code-may%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












      $begingroup$

      Here is a possible solution using the Principle of Inclusion/Exclusion:



      $textbf{Case 1}$ (digit is repeated $3$ times):



      Choose the repeated digit in $10 choose 1$ ways.



      Choose the remaining digit in $9 choose 1$ ways.



      Order the digits in $frac{4!}{3!} = 4$ ways.



      $textbf{Case 2}$ (digit is repeated $4$ times):



      Choose the repeated digit in $10 choose 1$ ways.



      Order the digits in $frac{4!}{4!} = 1$ way.



      So the number of combinations that don't satisfy your constraint is $(10 * 9 * 4) + (10 * 1) = 370$. Subtracting these from the $10,000$ total combinations yields $textbf{9,630}$ ways.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Here is a possible solution using the Principle of Inclusion/Exclusion:



        $textbf{Case 1}$ (digit is repeated $3$ times):



        Choose the repeated digit in $10 choose 1$ ways.



        Choose the remaining digit in $9 choose 1$ ways.



        Order the digits in $frac{4!}{3!} = 4$ ways.



        $textbf{Case 2}$ (digit is repeated $4$ times):



        Choose the repeated digit in $10 choose 1$ ways.



        Order the digits in $frac{4!}{4!} = 1$ way.



        So the number of combinations that don't satisfy your constraint is $(10 * 9 * 4) + (10 * 1) = 370$. Subtracting these from the $10,000$ total combinations yields $textbf{9,630}$ ways.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Here is a possible solution using the Principle of Inclusion/Exclusion:



          $textbf{Case 1}$ (digit is repeated $3$ times):



          Choose the repeated digit in $10 choose 1$ ways.



          Choose the remaining digit in $9 choose 1$ ways.



          Order the digits in $frac{4!}{3!} = 4$ ways.



          $textbf{Case 2}$ (digit is repeated $4$ times):



          Choose the repeated digit in $10 choose 1$ ways.



          Order the digits in $frac{4!}{4!} = 1$ way.



          So the number of combinations that don't satisfy your constraint is $(10 * 9 * 4) + (10 * 1) = 370$. Subtracting these from the $10,000$ total combinations yields $textbf{9,630}$ ways.






          share|cite|improve this answer











          $endgroup$



          Here is a possible solution using the Principle of Inclusion/Exclusion:



          $textbf{Case 1}$ (digit is repeated $3$ times):



          Choose the repeated digit in $10 choose 1$ ways.



          Choose the remaining digit in $9 choose 1$ ways.



          Order the digits in $frac{4!}{3!} = 4$ ways.



          $textbf{Case 2}$ (digit is repeated $4$ times):



          Choose the repeated digit in $10 choose 1$ ways.



          Order the digits in $frac{4!}{4!} = 1$ way.



          So the number of combinations that don't satisfy your constraint is $(10 * 9 * 4) + (10 * 1) = 370$. Subtracting these from the $10,000$ total combinations yields $textbf{9,630}$ ways.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 1:19

























          answered Dec 14 '18 at 0:51









          wjmcalliwjmcalli

          5115




          5115























              1












              $begingroup$

              Solution without using the Principle of Inclusion/Exclusion:



              Add by number of pairs of numbers: $0$ pairs $+$ $1$ pair $+$ $2$ pairs



              $=$ (pick $4$ of the $10$ order matters) $+$
              (pick which will be repeated and where and the other $2$ numbers order matters) $+$
              (pick number for each pair and location of first and second pair)



              $=P(10,4) + 10*{{4}choose{2}}*P(9,2) + {{10}choose{2}}*{{4}choose{2}}*1$



              $=5040$ $+$ $4320$ $+$ $290$



              $= 9630$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Solution without using the Principle of Inclusion/Exclusion:



                Add by number of pairs of numbers: $0$ pairs $+$ $1$ pair $+$ $2$ pairs



                $=$ (pick $4$ of the $10$ order matters) $+$
                (pick which will be repeated and where and the other $2$ numbers order matters) $+$
                (pick number for each pair and location of first and second pair)



                $=P(10,4) + 10*{{4}choose{2}}*P(9,2) + {{10}choose{2}}*{{4}choose{2}}*1$



                $=5040$ $+$ $4320$ $+$ $290$



                $= 9630$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Solution without using the Principle of Inclusion/Exclusion:



                  Add by number of pairs of numbers: $0$ pairs $+$ $1$ pair $+$ $2$ pairs



                  $=$ (pick $4$ of the $10$ order matters) $+$
                  (pick which will be repeated and where and the other $2$ numbers order matters) $+$
                  (pick number for each pair and location of first and second pair)



                  $=P(10,4) + 10*{{4}choose{2}}*P(9,2) + {{10}choose{2}}*{{4}choose{2}}*1$



                  $=5040$ $+$ $4320$ $+$ $290$



                  $= 9630$






                  share|cite|improve this answer









                  $endgroup$



                  Solution without using the Principle of Inclusion/Exclusion:



                  Add by number of pairs of numbers: $0$ pairs $+$ $1$ pair $+$ $2$ pairs



                  $=$ (pick $4$ of the $10$ order matters) $+$
                  (pick which will be repeated and where and the other $2$ numbers order matters) $+$
                  (pick number for each pair and location of first and second pair)



                  $=P(10,4) + 10*{{4}choose{2}}*P(9,2) + {{10}choose{2}}*{{4}choose{2}}*1$



                  $=5040$ $+$ $4320$ $+$ $290$



                  $= 9630$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 17:37









                  miniparserminiparser

                  7671616




                  7671616






























                      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%2f3038773%2fhow-many-4-digit-lock-combinations-are-possible-if-each-digit-in-the-code-may%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

                      Bressuire

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                      Read more

                      Gyllenstierna

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more