How many ways are there to divide $n$ dancers into dance circles?
$begingroup$
We have $n$ dancers, and they need to dance in circles. The order of the dancers inside each circle does matter. The circles themselves aren't ordered.
In the original question there was an additional requirement that the size of each dance circle is at least $2$, but I think that part can be completed later using the Inclusion–exclusion principle.
combinatorics
edited Dec 2 '18 at 11:10
Henrik
6,03792030
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
$endgroup$
add a comment |
$begingroup$
We have $n$ dancers, and they need to dance in circles. The order of the dancers inside each circle does matter. The circles themselves aren't ordered.
In the original question there was an additional requirement that the size of each dance circle is at least $2$, but I think that part can be completed later using the Inclusion–exclusion principle.
combinatorics
edited Dec 2 '18 at 11:10
Henrik
6,03792030
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
$endgroup$
$begingroup$
Have you calculated the answer for small values of $n$?
$endgroup$
– Gerry Myerson
Dec 2 '18 at 11:33
$begingroup$
I haven't, but since the answer turned out to be so simple, I assume that would have helped my figure it out. Though if I would have tried small n, I would have included the second restriction, which makes the formula more complex.
$endgroup$
– Amit Levy
Dec 3 '18 at 6:57
add a comment |
$begingroup$
We have $n$ dancers, and they need to dance in circles. The order of the dancers inside each circle does matter. The circles themselves aren't ordered.
In the original question there was an additional requirement that the size of each dance circle is at least $2$, but I think that part can be completed later using the Inclusion–exclusion principle.
combinatorics
edited Dec 2 '18 at 11:10
Henrik
6,03792030
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
$endgroup$
We have $n$ dancers, and they need to dance in circles. The order of the dancers inside each circle does matter. The circles themselves aren't ordered.
In the original question there was an additional requirement that the size of each dance circle is at least $2$, but I think that part can be completed later using the Inclusion–exclusion principle.
combinatorics
combinatorics
edited Dec 2 '18 at 11:10
Henrik
6,03792030
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
edited Dec 2 '18 at 11:10
Henrik
6,03792030
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
edited Dec 2 '18 at 11:10
Henrik
6,03792030
edited Dec 2 '18 at 11:10
Henrik
6,03792030
edited Dec 2 '18 at 11:10
Henrik
6,03792030
6,03792030
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
asked Dec 2 '18 at 11:08
Amit LevyAmit Levy
747
747
$begingroup$
Have you calculated the answer for small values of $n$?
$endgroup$
– Gerry Myerson
Dec 2 '18 at 11:33
$begingroup$
I haven't, but since the answer turned out to be so simple, I assume that would have helped my figure it out. Though if I would have tried small n, I would have included the second restriction, which makes the formula more complex.
$endgroup$
– Amit Levy
Dec 3 '18 at 6:57
add a comment |
$begingroup$
Have you calculated the answer for small values of $n$?
$endgroup$
– Gerry Myerson
Dec 2 '18 at 11:33
$begingroup$
I haven't, but since the answer turned out to be so simple, I assume that would have helped my figure it out. Though if I would have tried small n, I would have included the second restriction, which makes the formula more complex.
$endgroup$
– Amit Levy
Dec 3 '18 at 6:57
$begingroup$
Have you calculated the answer for small values of $n$?
$endgroup$
– Gerry Myerson
Dec 2 '18 at 11:33
$begingroup$
Have you calculated the answer for small values of $n$?
$endgroup$
– Gerry Myerson
Dec 2 '18 at 11:33
$begingroup$
I haven't, but since the answer turned out to be so simple, I assume that would have helped my figure it out. Though if I would have tried small n, I would have included the second restriction, which makes the formula more complex.
$endgroup$
– Amit Levy
Dec 3 '18 at 6:57
$begingroup$
I haven't, but since the answer turned out to be so simple, I assume that would have helped my figure it out. Though if I would have tried small n, I would have included the second restriction, which makes the formula more complex.
$endgroup$
– Amit Levy
Dec 3 '18 at 6:57
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Suppose we have 3 dancers, then the possible divisions are:
$$S_3={ (1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)}$$
The set $S_n$ is the set of permutations of $n$ elements. The permutations are written in cycle notation. Each cycle happens to correspond exactly with a circle of dancers.
In other words, the number of divisions of $n$ dancers is the number of elements in $S_n$, which is $n!$.
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Suppose we have 3 dancers, then the possible divisions are:
$$S_3={ (1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)}$$
The set $S_n$ is the set of permutations of $n$ elements. The permutations are written in cycle notation. Each cycle happens to correspond exactly with a circle of dancers.
In other words, the number of divisions of $n$ dancers is the number of elements in $S_n$, which is $n!$.
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
$endgroup$
add a comment |
$begingroup$
Suppose we have 3 dancers, then the possible divisions are:
$$S_3={ (1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)}$$
The set $S_n$ is the set of permutations of $n$ elements. The permutations are written in cycle notation. Each cycle happens to correspond exactly with a circle of dancers.
In other words, the number of divisions of $n$ dancers is the number of elements in $S_n$, which is $n!$.
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
$endgroup$
add a comment |
$begingroup$
Suppose we have 3 dancers, then the possible divisions are:
$$S_3={ (1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)}$$
The set $S_n$ is the set of permutations of $n$ elements. The permutations are written in cycle notation. Each cycle happens to correspond exactly with a circle of dancers.
In other words, the number of divisions of $n$ dancers is the number of elements in $S_n$, which is $n!$.
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
$endgroup$
Suppose we have 3 dancers, then the possible divisions are:
$$S_3={ (1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)}$$
The set $S_n$ is the set of permutations of $n$ elements. The permutations are written in cycle notation. Each cycle happens to correspond exactly with a circle of dancers.
In other words, the number of divisions of $n$ dancers is the number of elements in $S_n$, which is $n!$.
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
answered Dec 2 '18 at 11:33
I like SerenaI like Serena
4,2221722
4,2221722
add a comment |
add a comment |
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$begingroup$
Have you calculated the answer for small values of $n$?
$endgroup$
– Gerry Myerson
Dec 2 '18 at 11:33
$begingroup$
I haven't, but since the answer turned out to be so simple, I assume that would have helped my figure it out. Though if I would have tried small n, I would have included the second restriction, which makes the formula more complex.
$endgroup$
– Amit Levy
Dec 3 '18 at 6:57