How to permutate numbers where repetition of immediate number is not allowed [closed]

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Hi i have been studying permutation and combination i fell into a problem.
if You have a set of letters to choose from = A,B,C,D, and you must choose 3 of them, but no immediate repetition of a letter.



Example A,B,A, C,D,C, etc is allowed.



What formular, how should i solve this problem?



Thanks in Advance




permutations combinations




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closed as off-topic by Henrik, John Ma, Strants, jgon, Taroccoesbrocco Jul 20 at 21:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, John Ma, Strants, jgon, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.
















    up vote
    0
    down vote

    favorite












    Hi i have been studying permutation and combination i fell into a problem.
    if You have a set of letters to choose from = A,B,C,D, and you must choose 3 of them, but no immediate repetition of a letter.



    Example A,B,A, C,D,C, etc is allowed.



    What formular, how should i solve this problem?



    Thanks in Advance




    permutations combinations




    share|cite|improve this question











    closed as off-topic by Henrik, John Ma, Strants, jgon, Taroccoesbrocco Jul 20 at 21:15


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, John Ma, Strants, jgon, Taroccoesbrocco
    If this question can be reworded to fit the rules in the help center, please edit the question.














      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Hi i have been studying permutation and combination i fell into a problem.
      if You have a set of letters to choose from = A,B,C,D, and you must choose 3 of them, but no immediate repetition of a letter.



      Example A,B,A, C,D,C, etc is allowed.



      What formular, how should i solve this problem?



      Thanks in Advance




      permutations combinations




      share|cite|improve this question











      Hi i have been studying permutation and combination i fell into a problem.
      if You have a set of letters to choose from = A,B,C,D, and you must choose 3 of them, but no immediate repetition of a letter.



      Example A,B,A, C,D,C, etc is allowed.



      What formular, how should i solve this problem?



      Thanks in Advance





      permutations combinations





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 20 at 13:54









      Declan

      32




      32




      closed as off-topic by Henrik, John Ma, Strants, jgon, Taroccoesbrocco Jul 20 at 21:15


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, John Ma, Strants, jgon, Taroccoesbrocco
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Henrik, John Ma, Strants, jgon, Taroccoesbrocco Jul 20 at 21:15


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, John Ma, Strants, jgon, Taroccoesbrocco
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
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          For the first item in the resulting set we have $4$ choices.



          For the second item in the set we have $4-1 = 3$ choices as we cannot repeat the previous letter.



          Finally for the third item in the set we have $4-1= 3$ choices as again we cannot repeat the previous letter.



          Multiplying the number of choices (using the rule of product) we have $4*3*3 = 36$ solutions.






          share|cite|improve this answer




























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            For the first item in the resulting set we have $4$ choices.



            For the second item in the set we have $4-1 = 3$ choices as we cannot repeat the previous letter.



            Finally for the third item in the set we have $4-1= 3$ choices as again we cannot repeat the previous letter.



            Multiplying the number of choices (using the rule of product) we have $4*3*3 = 36$ solutions.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              For the first item in the resulting set we have $4$ choices.



              For the second item in the set we have $4-1 = 3$ choices as we cannot repeat the previous letter.



              Finally for the third item in the set we have $4-1= 3$ choices as again we cannot repeat the previous letter.



              Multiplying the number of choices (using the rule of product) we have $4*3*3 = 36$ solutions.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                For the first item in the resulting set we have $4$ choices.



                For the second item in the set we have $4-1 = 3$ choices as we cannot repeat the previous letter.



                Finally for the third item in the set we have $4-1= 3$ choices as again we cannot repeat the previous letter.



                Multiplying the number of choices (using the rule of product) we have $4*3*3 = 36$ solutions.






                share|cite|improve this answer













                For the first item in the resulting set we have $4$ choices.



                For the second item in the set we have $4-1 = 3$ choices as we cannot repeat the previous letter.



                Finally for the third item in the set we have $4-1= 3$ choices as again we cannot repeat the previous letter.



                Multiplying the number of choices (using the rule of product) we have $4*3*3 = 36$ solutions.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 20 at 14:02









                packetpacket

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