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









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      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






























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