In how many ways can 3 employees visit 40 locations
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Three employees need to visit 40 different cities under the following conditions: each location should be visited by exactly one employee, and no location should be visited multiple times. The travel agency should plan the tours for three employees. How many "tour" options are there in total?
Is the solution number of permutations with 3 cycles?
combinatorics permutation-cycles
asked Jan 14 at 14:26
StratocarterStratocarter
126
$endgroup$
add a comment |
$begingroup$
Three employees need to visit 40 different cities under the following conditions: each location should be visited by exactly one employee, and no location should be visited multiple times. The travel agency should plan the tours for three employees. How many "tour" options are there in total?
Is the solution number of permutations with 3 cycles?
combinatorics permutation-cycles
asked Jan 14 at 14:26
StratocarterStratocarter
126
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$begingroup$
Could you edit to make a precise definition of what is distinguishable and what is not ?
$endgroup$
– P. Quinton
Jan 14 at 14:54
add a comment |
$begingroup$
Three employees need to visit 40 different cities under the following conditions: each location should be visited by exactly one employee, and no location should be visited multiple times. The travel agency should plan the tours for three employees. How many "tour" options are there in total?
Is the solution number of permutations with 3 cycles?
combinatorics permutation-cycles
asked Jan 14 at 14:26
StratocarterStratocarter
126
$endgroup$
Three employees need to visit 40 different cities under the following conditions: each location should be visited by exactly one employee, and no location should be visited multiple times. The travel agency should plan the tours for three employees. How many "tour" options are there in total?
Is the solution number of permutations with 3 cycles?
combinatorics permutation-cycles
combinatorics permutation-cycles
asked Jan 14 at 14:26
StratocarterStratocarter
126
asked Jan 14 at 14:26
StratocarterStratocarter
126
edited Jan 15 at 12:22
Stratocarter
asked Jan 14 at 14:26
StratocarterStratocarter
126
asked Jan 14 at 14:26
StratocarterStratocarter
126
asked Jan 14 at 14:26
StratocarterStratocarter
126
126
$begingroup$
Could you edit to make a precise definition of what is distinguishable and what is not ?
$endgroup$
– P. Quinton
Jan 14 at 14:54
add a comment |
$begingroup$
Could you edit to make a precise definition of what is distinguishable and what is not ?
$endgroup$
– P. Quinton
Jan 14 at 14:54
$begingroup$
Could you edit to make a precise definition of what is distinguishable and what is not ?
$endgroup$
– P. Quinton
Jan 14 at 14:54
$begingroup$
Could you edit to make a precise definition of what is distinguishable and what is not ?
$endgroup$
– P. Quinton
Jan 14 at 14:54
add a comment |
2 Answers
2
active
oldest
votes
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For each location you have 3 chooises (employees)and you have 40 locations so $3^{40}$?
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
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$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
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No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
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Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
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Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
add a comment |
$begingroup$
It seems that you want to partition the given set of $40$ distinguishable locations into $3$ nonempty blocks, without naming the blocks. The number of ways this can be done is the Stirling number of the second kind $S(40,3)$ (there are various notations in use). The resulting number is
$$S(40,3)=2,026,277,026,753,674,246 .$$
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
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$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For each location you have 3 chooises (employees)and you have 40 locations so $3^{40}$?
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
$endgroup$
$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
$begingroup$
No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
$begingroup$
Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
$begingroup$
Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
add a comment |
$begingroup$
For each location you have 3 chooises (employees)and you have 40 locations so $3^{40}$?
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
$endgroup$
$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
$begingroup$
No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
$begingroup$
Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
$begingroup$
Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
add a comment |
$begingroup$
For each location you have 3 chooises (employees)and you have 40 locations so $3^{40}$?
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
$endgroup$
For each location you have 3 chooises (employees)and you have 40 locations so $3^{40}$?
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
answered Jan 14 at 14:30
greedoidgreedoid
40.7k1149100
40.7k1149100
$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
$begingroup$
No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
$begingroup$
Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
$begingroup$
Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
add a comment |
$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
$begingroup$
No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
$begingroup$
Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
$begingroup$
Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
$begingroup$
Are the three employees distinguishable ? do we double count assignment $A,B,C$ and $A,C,B$ for instance ?
$endgroup$
– P. Quinton
Jan 14 at 14:44
$begingroup$
No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
$begingroup$
No, they are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 14:46
$begingroup$
Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
$begingroup$
Employees are not distinguishable? How come?
$endgroup$
– greedoid
Jan 14 at 14:48
$begingroup$
Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
$begingroup$
Employees are the same, like 3 stars.
$endgroup$
– Stratocarter
Jan 14 at 14:52
add a comment |
$begingroup$
It seems that you want to partition the given set of $40$ distinguishable locations into $3$ nonempty blocks, without naming the blocks. The number of ways this can be done is the Stirling number of the second kind $S(40,3)$ (there are various notations in use). The resulting number is
$$S(40,3)=2,026,277,026,753,674,246 .$$
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
add a comment |
$begingroup$
It seems that you want to partition the given set of $40$ distinguishable locations into $3$ nonempty blocks, without naming the blocks. The number of ways this can be done is the Stirling number of the second kind $S(40,3)$ (there are various notations in use). The resulting number is
$$S(40,3)=2,026,277,026,753,674,246 .$$
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
add a comment |
$begingroup$
It seems that you want to partition the given set of $40$ distinguishable locations into $3$ nonempty blocks, without naming the blocks. The number of ways this can be done is the Stirling number of the second kind $S(40,3)$ (there are various notations in use). The resulting number is
$$S(40,3)=2,026,277,026,753,674,246 .$$
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
$endgroup$
It seems that you want to partition the given set of $40$ distinguishable locations into $3$ nonempty blocks, without naming the blocks. The number of ways this can be done is the Stirling number of the second kind $S(40,3)$ (there are various notations in use). The resulting number is
$$S(40,3)=2,026,277,026,753,674,246 .$$
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
answered Jan 14 at 16:02
Christian BlatterChristian Blatter
173k7113326
173k7113326
$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
add a comment |
$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
$begingroup$
Thank you, now I see that I wasn't clear. I've edited my question.
$endgroup$
– Stratocarter
Jan 14 at 16:13
add a comment |
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$begingroup$
Could you edit to make a precise definition of what is distinguishable and what is not ?
$endgroup$
– P. Quinton
Jan 14 at 14:54