How to apply the principle of inclusion and exclusion in this problem

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8 new employees are assigned to 3 different departments. How many differents ways of assigning the employees exist if each department has to receive at least one employee?



At first I thought to work with the departments as sets but I think that is not possible beacuse the problem does not say that the employees can work in one or more departmets. Is working with the departments as sets the right option?







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  • see this wanswer also: math.stackexchange.com/questions/550256/…
    – Hector Blandin
    Jul 31 at 4:00














up vote
4
down vote

favorite












8 new employees are assigned to 3 different departments. How many differents ways of assigning the employees exist if each department has to receive at least one employee?



At first I thought to work with the departments as sets but I think that is not possible beacuse the problem does not say that the employees can work in one or more departmets. Is working with the departments as sets the right option?







share|cite|improve this question



















  • see this wanswer also: math.stackexchange.com/questions/550256/…
    – Hector Blandin
    Jul 31 at 4:00












up vote
4
down vote

favorite









up vote
4
down vote

favorite











8 new employees are assigned to 3 different departments. How many differents ways of assigning the employees exist if each department has to receive at least one employee?



At first I thought to work with the departments as sets but I think that is not possible beacuse the problem does not say that the employees can work in one or more departmets. Is working with the departments as sets the right option?







share|cite|improve this question











8 new employees are assigned to 3 different departments. How many differents ways of assigning the employees exist if each department has to receive at least one employee?



At first I thought to work with the departments as sets but I think that is not possible beacuse the problem does not say that the employees can work in one or more departmets. Is working with the departments as sets the right option?









share|cite|improve this question










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asked Jul 31 at 2:56









Jorge Chang

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  • see this wanswer also: math.stackexchange.com/questions/550256/…
    – Hector Blandin
    Jul 31 at 4:00
















  • see this wanswer also: math.stackexchange.com/questions/550256/…
    – Hector Blandin
    Jul 31 at 4:00















see this wanswer also: math.stackexchange.com/questions/550256/…
– Hector Blandin
Jul 31 at 4:00




see this wanswer also: math.stackexchange.com/questions/550256/…
– Hector Blandin
Jul 31 at 4:00










2 Answers
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The number of ways to map $8$ people into $3$ departments is $N(0)=3^8$. The number of ways to miss a particular department would be $binom322^8$. The number of ways to miss two particular departments would be $binom311^8$. The number of ways to miss three particular departments would be $binom300^8$.



Inclusion-Exclusion says there are
$$
3^8-binom322^8+binom311^8-binom300^8=5796
$$
ways to distribute $8$ people among $3$ departments with no empty departments.






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    up vote
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    down vote













    This is equivalent to count the number of surjective maps from the set of employees $e_1,dots,e_8$ to the set of departments $d_1,d_2,d_3$. So you get
    $$S(8,3)times 3!=5796quad textways$$
    where $S(8,3)$ is the Stirling number of second kind.
    Recall that the condition each department has to receive at least one employee means that each map has to be surjective.




    The number of surjective maps from an $n$ elements set onto a $k$ elements set is given by:
    $$S(n,k)cdot k!$$







    share|cite|improve this answer























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      2 Answers
      2






      active

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      2 Answers
      2






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      active

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      up vote
      4
      down vote



      accepted










      The number of ways to map $8$ people into $3$ departments is $N(0)=3^8$. The number of ways to miss a particular department would be $binom322^8$. The number of ways to miss two particular departments would be $binom311^8$. The number of ways to miss three particular departments would be $binom300^8$.



      Inclusion-Exclusion says there are
      $$
      3^8-binom322^8+binom311^8-binom300^8=5796
      $$
      ways to distribute $8$ people among $3$ departments with no empty departments.






      share|cite|improve this answer

























        up vote
        4
        down vote



        accepted










        The number of ways to map $8$ people into $3$ departments is $N(0)=3^8$. The number of ways to miss a particular department would be $binom322^8$. The number of ways to miss two particular departments would be $binom311^8$. The number of ways to miss three particular departments would be $binom300^8$.



        Inclusion-Exclusion says there are
        $$
        3^8-binom322^8+binom311^8-binom300^8=5796
        $$
        ways to distribute $8$ people among $3$ departments with no empty departments.






        share|cite|improve this answer























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          The number of ways to map $8$ people into $3$ departments is $N(0)=3^8$. The number of ways to miss a particular department would be $binom322^8$. The number of ways to miss two particular departments would be $binom311^8$. The number of ways to miss three particular departments would be $binom300^8$.



          Inclusion-Exclusion says there are
          $$
          3^8-binom322^8+binom311^8-binom300^8=5796
          $$
          ways to distribute $8$ people among $3$ departments with no empty departments.






          share|cite|improve this answer













          The number of ways to map $8$ people into $3$ departments is $N(0)=3^8$. The number of ways to miss a particular department would be $binom322^8$. The number of ways to miss two particular departments would be $binom311^8$. The number of ways to miss three particular departments would be $binom300^8$.



          Inclusion-Exclusion says there are
          $$
          3^8-binom322^8+binom311^8-binom300^8=5796
          $$
          ways to distribute $8$ people among $3$ departments with no empty departments.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 31 at 3:54









          robjohn♦

          258k25296612




          258k25296612




















              up vote
              1
              down vote













              This is equivalent to count the number of surjective maps from the set of employees $e_1,dots,e_8$ to the set of departments $d_1,d_2,d_3$. So you get
              $$S(8,3)times 3!=5796quad textways$$
              where $S(8,3)$ is the Stirling number of second kind.
              Recall that the condition each department has to receive at least one employee means that each map has to be surjective.




              The number of surjective maps from an $n$ elements set onto a $k$ elements set is given by:
              $$S(n,k)cdot k!$$







              share|cite|improve this answer



























                up vote
                1
                down vote













                This is equivalent to count the number of surjective maps from the set of employees $e_1,dots,e_8$ to the set of departments $d_1,d_2,d_3$. So you get
                $$S(8,3)times 3!=5796quad textways$$
                where $S(8,3)$ is the Stirling number of second kind.
                Recall that the condition each department has to receive at least one employee means that each map has to be surjective.




                The number of surjective maps from an $n$ elements set onto a $k$ elements set is given by:
                $$S(n,k)cdot k!$$







                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  This is equivalent to count the number of surjective maps from the set of employees $e_1,dots,e_8$ to the set of departments $d_1,d_2,d_3$. So you get
                  $$S(8,3)times 3!=5796quad textways$$
                  where $S(8,3)$ is the Stirling number of second kind.
                  Recall that the condition each department has to receive at least one employee means that each map has to be surjective.




                  The number of surjective maps from an $n$ elements set onto a $k$ elements set is given by:
                  $$S(n,k)cdot k!$$







                  share|cite|improve this answer















                  This is equivalent to count the number of surjective maps from the set of employees $e_1,dots,e_8$ to the set of departments $d_1,d_2,d_3$. So you get
                  $$S(8,3)times 3!=5796quad textways$$
                  where $S(8,3)$ is the Stirling number of second kind.
                  Recall that the condition each department has to receive at least one employee means that each map has to be surjective.




                  The number of surjective maps from an $n$ elements set onto a $k$ elements set is given by:
                  $$S(n,k)cdot k!$$








                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 31 at 4:03


























                  answered Jul 31 at 3:20









                  Hector Blandin

                  1,741716




                  1,741716






















                       

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