How many ways are there to put 4 distinguishable balls into 2 indistinguishable boxes? [closed]












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I do not know the answer! How do you do this type of question? Please type the answer and show me the steps to solve it! Thanks






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closed as off-topic by Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici Jan 29 at 9:29


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$


    I do not know the answer! How do you do this type of question? Please type the answer and show me the steps to solve it! Thanks






    combinatorics







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici Jan 29 at 9:29


    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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0


      0



      $begingroup$


      I do not know the answer! How do you do this type of question? Please type the answer and show me the steps to solve it! Thanks






      combinatorics







      share|cite|improve this question











      $endgroup$




      I do not know the answer! How do you do this type of question? Please type the answer and show me the steps to solve it! Thanks






      combinatorics




      combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 0:47









      WaveX

      2,7642822




      2,7642822










      asked Jan 28 at 0:35









      I'm not a masterI'm not a master

      32




      32




      closed as off-topic by Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici Jan 29 at 9:29


      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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici Jan 29 at 9:29


      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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Lord Shark the Unknown, Leucippus, José Carlos Santos, Claude Leibovici

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
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          $begingroup$

          For these small problems, you can approach directly via brute force or smart counting.



          Let our balls be labeled $A,B,C,D$.



          We know that ball $A$ will be in some box... but we don't care which box it is since we can't tell them apart. Now that ball $A$ is in that box however, we can now tell the two boxes apart. The box with $A$ in it, and the box without $A$ in it.



          For each of the balls $B,C,D$, choose whether they go in the box with $A$ or into the box without $A$. Application of the rule of product/multiplication principle tells us then that there are $2times 2times 2 = 8$ such outcomes.




          For the more general problem of having $n$ distinguishable balls and $k$ indistinguishable boxes, this will rely on stirling numbers of the second kind. $left{begin{smallmatrix}n\rend{smallmatrix}right}$ counts the number of ways to partition $n$ distinguishable objects into $r$ otherwise indistinguishable non-empty sets. Assuming you allow some boxes to be empty, we get as a result by summing over all possible number of non-empty boxes a total of:



          $$sumlimits_{r=0}^kleft{begin{matrix}n\rend{matrix}right}$$






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            0












            $begingroup$

            Ball $1$ has $2$ options, ball $2$ has $2$ options, $dots$



            This will give us how many total options? Note that some of these options are actually the same, since our boxes are indistinguishable, so we will need to figure out how much we are over-counting by.





            Another way to look at it is this way: think of it as pairing the numbers $1,2,3,4$ together. How many ways can we pair up all of them together? How many ways can we pair them up so that we have $3$ numbers in one group and a lone number by itself? Then how many ways can we pair them up with $2$ in one group and $2$ in the other?




            share|cite|improve this answer









            $endgroup$




















              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              For these small problems, you can approach directly via brute force or smart counting.



              Let our balls be labeled $A,B,C,D$.



              We know that ball $A$ will be in some box... but we don't care which box it is since we can't tell them apart. Now that ball $A$ is in that box however, we can now tell the two boxes apart. The box with $A$ in it, and the box without $A$ in it.



              For each of the balls $B,C,D$, choose whether they go in the box with $A$ or into the box without $A$. Application of the rule of product/multiplication principle tells us then that there are $2times 2times 2 = 8$ such outcomes.




              For the more general problem of having $n$ distinguishable balls and $k$ indistinguishable boxes, this will rely on stirling numbers of the second kind. $left{begin{smallmatrix}n\rend{smallmatrix}right}$ counts the number of ways to partition $n$ distinguishable objects into $r$ otherwise indistinguishable non-empty sets. Assuming you allow some boxes to be empty, we get as a result by summing over all possible number of non-empty boxes a total of:



              $$sumlimits_{r=0}^kleft{begin{matrix}n\rend{matrix}right}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For these small problems, you can approach directly via brute force or smart counting.



                Let our balls be labeled $A,B,C,D$.



                We know that ball $A$ will be in some box... but we don't care which box it is since we can't tell them apart. Now that ball $A$ is in that box however, we can now tell the two boxes apart. The box with $A$ in it, and the box without $A$ in it.



                For each of the balls $B,C,D$, choose whether they go in the box with $A$ or into the box without $A$. Application of the rule of product/multiplication principle tells us then that there are $2times 2times 2 = 8$ such outcomes.




                For the more general problem of having $n$ distinguishable balls and $k$ indistinguishable boxes, this will rely on stirling numbers of the second kind. $left{begin{smallmatrix}n\rend{smallmatrix}right}$ counts the number of ways to partition $n$ distinguishable objects into $r$ otherwise indistinguishable non-empty sets. Assuming you allow some boxes to be empty, we get as a result by summing over all possible number of non-empty boxes a total of:



                $$sumlimits_{r=0}^kleft{begin{matrix}n\rend{matrix}right}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For these small problems, you can approach directly via brute force or smart counting.



                  Let our balls be labeled $A,B,C,D$.



                  We know that ball $A$ will be in some box... but we don't care which box it is since we can't tell them apart. Now that ball $A$ is in that box however, we can now tell the two boxes apart. The box with $A$ in it, and the box without $A$ in it.



                  For each of the balls $B,C,D$, choose whether they go in the box with $A$ or into the box without $A$. Application of the rule of product/multiplication principle tells us then that there are $2times 2times 2 = 8$ such outcomes.




                  For the more general problem of having $n$ distinguishable balls and $k$ indistinguishable boxes, this will rely on stirling numbers of the second kind. $left{begin{smallmatrix}n\rend{smallmatrix}right}$ counts the number of ways to partition $n$ distinguishable objects into $r$ otherwise indistinguishable non-empty sets. Assuming you allow some boxes to be empty, we get as a result by summing over all possible number of non-empty boxes a total of:



                  $$sumlimits_{r=0}^kleft{begin{matrix}n\rend{matrix}right}$$






                  share|cite|improve this answer









                  $endgroup$



                  For these small problems, you can approach directly via brute force or smart counting.



                  Let our balls be labeled $A,B,C,D$.



                  We know that ball $A$ will be in some box... but we don't care which box it is since we can't tell them apart. Now that ball $A$ is in that box however, we can now tell the two boxes apart. The box with $A$ in it, and the box without $A$ in it.



                  For each of the balls $B,C,D$, choose whether they go in the box with $A$ or into the box without $A$. Application of the rule of product/multiplication principle tells us then that there are $2times 2times 2 = 8$ such outcomes.




                  For the more general problem of having $n$ distinguishable balls and $k$ indistinguishable boxes, this will rely on stirling numbers of the second kind. $left{begin{smallmatrix}n\rend{smallmatrix}right}$ counts the number of ways to partition $n$ distinguishable objects into $r$ otherwise indistinguishable non-empty sets. Assuming you allow some boxes to be empty, we get as a result by summing over all possible number of non-empty boxes a total of:



                  $$sumlimits_{r=0}^kleft{begin{matrix}n\rend{matrix}right}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 28 at 0:47









                  JMoravitzJMoravitz

                  48.7k43988




                  48.7k43988























                      0












                      $begingroup$

                      Ball $1$ has $2$ options, ball $2$ has $2$ options, $dots$



                      This will give us how many total options? Note that some of these options are actually the same, since our boxes are indistinguishable, so we will need to figure out how much we are over-counting by.





                      Another way to look at it is this way: think of it as pairing the numbers $1,2,3,4$ together. How many ways can we pair up all of them together? How many ways can we pair them up so that we have $3$ numbers in one group and a lone number by itself? Then how many ways can we pair them up with $2$ in one group and $2$ in the other?




                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Ball $1$ has $2$ options, ball $2$ has $2$ options, $dots$



                        This will give us how many total options? Note that some of these options are actually the same, since our boxes are indistinguishable, so we will need to figure out how much we are over-counting by.





                        Another way to look at it is this way: think of it as pairing the numbers $1,2,3,4$ together. How many ways can we pair up all of them together? How many ways can we pair them up so that we have $3$ numbers in one group and a lone number by itself? Then how many ways can we pair them up with $2$ in one group and $2$ in the other?




                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Ball $1$ has $2$ options, ball $2$ has $2$ options, $dots$



                          This will give us how many total options? Note that some of these options are actually the same, since our boxes are indistinguishable, so we will need to figure out how much we are over-counting by.





                          Another way to look at it is this way: think of it as pairing the numbers $1,2,3,4$ together. How many ways can we pair up all of them together? How many ways can we pair them up so that we have $3$ numbers in one group and a lone number by itself? Then how many ways can we pair them up with $2$ in one group and $2$ in the other?




                          share|cite|improve this answer









                          $endgroup$



                          Ball $1$ has $2$ options, ball $2$ has $2$ options, $dots$



                          This will give us how many total options? Note that some of these options are actually the same, since our boxes are indistinguishable, so we will need to figure out how much we are over-counting by.





                          Another way to look at it is this way: think of it as pairing the numbers $1,2,3,4$ together. How many ways can we pair up all of them together? How many ways can we pair them up so that we have $3$ numbers in one group and a lone number by itself? Then how many ways can we pair them up with $2$ in one group and $2$ in the other?





                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 28 at 0:45









                          WaveXWaveX

                          2,7642822




                          2,7642822















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