Probability 12 families meeting

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please help me to find the solution of this.



What is the Probability that 12 families meet 4 times a year an every time only 4 families meeting together without repetition so that each family sees the other family only once without repetition.



For example at the first time will be



1 2 3 4



5 6 7 8



9 10 11 12



What is the 4th probability?



At the second time



3 6 9 12



That is wrong because the families 9 & 12 should not be together another time



I am waiting for the answer
Best regards




probability




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  • Could you please clarify a bit or post the question as written? It's a little hard to understand what you're asking here.
    – CaptainAmerica16
    Aug 2 at 7:38














up vote
0
down vote

favorite












please help me to find the solution of this.



What is the Probability that 12 families meet 4 times a year an every time only 4 families meeting together without repetition so that each family sees the other family only once without repetition.



For example at the first time will be



1 2 3 4



5 6 7 8



9 10 11 12



What is the 4th probability?



At the second time



3 6 9 12



That is wrong because the families 9 & 12 should not be together another time



I am waiting for the answer
Best regards




probability




share|cite|improve this question





















  • Could you please clarify a bit or post the question as written? It's a little hard to understand what you're asking here.
    – CaptainAmerica16
    Aug 2 at 7:38












up vote
0
down vote

favorite









up vote
0
down vote

favorite











please help me to find the solution of this.



What is the Probability that 12 families meet 4 times a year an every time only 4 families meeting together without repetition so that each family sees the other family only once without repetition.



For example at the first time will be



1 2 3 4



5 6 7 8



9 10 11 12



What is the 4th probability?



At the second time



3 6 9 12



That is wrong because the families 9 & 12 should not be together another time



I am waiting for the answer
Best regards




probability




share|cite|improve this question













please help me to find the solution of this.



What is the Probability that 12 families meet 4 times a year an every time only 4 families meeting together without repetition so that each family sees the other family only once without repetition.



For example at the first time will be



1 2 3 4



5 6 7 8



9 10 11 12



What is the 4th probability?



At the second time



3 6 9 12



That is wrong because the families 9 & 12 should not be together another time



I am waiting for the answer
Best regards





probability





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 7:53









Henry

92.7k469147




92.7k469147









asked Aug 2 at 7:01









Lady

1




1











  • Could you please clarify a bit or post the question as written? It's a little hard to understand what you're asking here.
    – CaptainAmerica16
    Aug 2 at 7:38
















  • Could you please clarify a bit or post the question as written? It's a little hard to understand what you're asking here.
    – CaptainAmerica16
    Aug 2 at 7:38















Could you please clarify a bit or post the question as written? It's a little hard to understand what you're asking here.
– CaptainAmerica16
Aug 2 at 7:38




Could you please clarify a bit or post the question as written? It's a little hard to understand what you're asking here.
– CaptainAmerica16
Aug 2 at 7:38










1 Answer
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It is not possible as you have structured it



At the second time, a family from the first group of the first time can only meet



  • one family from the second group of the first time

  • one family from the third group of the first time

  • leaving a fourth possibility who cannot be from any of the three groups of the first time

(Perhaps there is only one group of four each time)






share|cite|improve this answer





















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    up vote
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    It is not possible as you have structured it



    At the second time, a family from the first group of the first time can only meet



    • one family from the second group of the first time

    • one family from the third group of the first time

    • leaving a fourth possibility who cannot be from any of the three groups of the first time

    (Perhaps there is only one group of four each time)






    share|cite|improve this answer

























      up vote
      0
      down vote













      It is not possible as you have structured it



      At the second time, a family from the first group of the first time can only meet



      • one family from the second group of the first time

      • one family from the third group of the first time

      • leaving a fourth possibility who cannot be from any of the three groups of the first time

      (Perhaps there is only one group of four each time)






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        It is not possible as you have structured it



        At the second time, a family from the first group of the first time can only meet



        • one family from the second group of the first time

        • one family from the third group of the first time

        • leaving a fourth possibility who cannot be from any of the three groups of the first time

        (Perhaps there is only one group of four each time)






        share|cite|improve this answer













        It is not possible as you have structured it



        At the second time, a family from the first group of the first time can only meet



        • one family from the second group of the first time

        • one family from the third group of the first time

        • leaving a fourth possibility who cannot be from any of the three groups of the first time

        (Perhaps there is only one group of four each time)







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 7:56









        Henry

        92.7k469147




        92.7k469147






















             

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