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?
combinatorics inclusion-exclusion
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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?
combinatorics inclusion-exclusion
see this wanswer also: math.stackexchange.com/questions/550256/…
– Hector Blandin
Jul 31 at 4:00
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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?
combinatorics inclusion-exclusion
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?
combinatorics inclusion-exclusion
asked Jul 31 at 2:56
Jorge Chang
254
254
see this wanswer also: math.stackexchange.com/questions/550256/…
– Hector Blandin
Jul 31 at 4:00
add a comment |Â
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
add a comment |Â
2 Answers
2
active
oldest
votes
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.
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1
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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:
$$S(n,k)cdot k!$$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |Â
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.
add a comment |Â
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.
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.
answered Jul 31 at 3:54
robjohn♦
258k25296612
258k25296612
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add a comment |Â
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:
$$S(n,k)cdot k!$$
add a comment |Â
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:
$$S(n,k)cdot k!$$
add a comment |Â
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:
$$S(n,k)cdot k!$$
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:
$$S(n,k)cdot k!$$
edited Jul 31 at 4:03
answered Jul 31 at 3:20


Hector Blandin
1,741716
1,741716
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see this wanswer also: math.stackexchange.com/questions/550256/…
– Hector Blandin
Jul 31 at 4:00