Combinatorics arranging repeated numbers











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In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?



I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?






combinatorics combinations







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    modified, sorry.
    – Rodrigo Pizarro
    Nov 17 at 23:59















up vote
1
down vote

favorite












In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?



I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?






combinatorics combinations







share|cite|improve this question




















  • 1




    modified, sorry.
    – Rodrigo Pizarro
    Nov 17 at 23:59













up vote
1
down vote

favorite









up vote
1
down vote

favorite











In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?



I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?






combinatorics combinations







share|cite|improve this question















In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?



I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?






combinatorics combinations




combinatorics combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 17 at 23:59

























asked Nov 17 at 23:51









Rodrigo Pizarro

834217




834217








  • 1




    modified, sorry.
    – Rodrigo Pizarro
    Nov 17 at 23:59














  • 1




    modified, sorry.
    – Rodrigo Pizarro
    Nov 17 at 23:59








1




1




modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59




modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59










1 Answer
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Your answer is correct.



Strategy: Consider cases, depending on the number of fives used.




  1. No fives are used: Choose three of the other three numbers in $B$.

  2. One five is used: Choose two of the other three numbers in $B$.

  3. Two fives are used: Choose one of the other three numbers in $B$.

  4. Three fives are used: Choose none of the other three numbers in $B$.


Since the cases are mutually exclusive and exhaustive, add the results to get the total.






share|cite|improve this answer





















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Your answer is correct.



    Strategy: Consider cases, depending on the number of fives used.




    1. No fives are used: Choose three of the other three numbers in $B$.

    2. One five is used: Choose two of the other three numbers in $B$.

    3. Two fives are used: Choose one of the other three numbers in $B$.

    4. Three fives are used: Choose none of the other three numbers in $B$.


    Since the cases are mutually exclusive and exhaustive, add the results to get the total.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Your answer is correct.



      Strategy: Consider cases, depending on the number of fives used.




      1. No fives are used: Choose three of the other three numbers in $B$.

      2. One five is used: Choose two of the other three numbers in $B$.

      3. Two fives are used: Choose one of the other three numbers in $B$.

      4. Three fives are used: Choose none of the other three numbers in $B$.


      Since the cases are mutually exclusive and exhaustive, add the results to get the total.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Your answer is correct.



        Strategy: Consider cases, depending on the number of fives used.




        1. No fives are used: Choose three of the other three numbers in $B$.

        2. One five is used: Choose two of the other three numbers in $B$.

        3. Two fives are used: Choose one of the other three numbers in $B$.

        4. Three fives are used: Choose none of the other three numbers in $B$.


        Since the cases are mutually exclusive and exhaustive, add the results to get the total.






        share|cite|improve this answer












        Your answer is correct.



        Strategy: Consider cases, depending on the number of fives used.




        1. No fives are used: Choose three of the other three numbers in $B$.

        2. One five is used: Choose two of the other three numbers in $B$.

        3. Two fives are used: Choose one of the other three numbers in $B$.

        4. Three fives are used: Choose none of the other three numbers in $B$.


        Since the cases are mutually exclusive and exhaustive, add the results to get the total.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 18 at 0:03









        N. F. Taussig

        42.8k93254




        42.8k93254






























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