How many ways are there to divide $n$ dancers into dance circles?
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 at 11:10
Henrik
5,95692030
asked Dec 2 at 11:08
Amit Levy
232
add a comment |
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 at 11:10
Henrik
5,95692030
asked Dec 2 at 11:08
Amit Levy
232
Have you calculated the answer for small values of $n$?
– Gerry Myerson
Dec 2 at 11:33
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.
– Amit Levy
Dec 3 at 6:57
add a comment |
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 at 11:10
Henrik
5,95692030
asked Dec 2 at 11:08
Amit Levy
232
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 at 11:10
Henrik
5,95692030
asked Dec 2 at 11:08
Amit Levy
232
edited Dec 2 at 11:10
Henrik
5,95692030
asked Dec 2 at 11:08
Amit Levy
232
edited Dec 2 at 11:10
Henrik
5,95692030
edited Dec 2 at 11:10
Henrik
5,95692030
edited Dec 2 at 11:10
Henrik
5,95692030
5,95692030
asked Dec 2 at 11:08
Amit Levy
232
asked Dec 2 at 11:08
Amit Levy
232
asked Dec 2 at 11:08
Amit Levy
232
232
Have you calculated the answer for small values of $n$?
– Gerry Myerson
Dec 2 at 11:33
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.
– Amit Levy
Dec 3 at 6:57
add a comment |
Have you calculated the answer for small values of $n$?
– Gerry Myerson
Dec 2 at 11:33
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.
– Amit Levy
Dec 3 at 6:57
Have you calculated the answer for small values of $n$?
– Gerry Myerson
Dec 2 at 11:33
Have you calculated the answer for small values of $n$?
– Gerry Myerson
Dec 2 at 11:33
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.
– Amit Levy
Dec 3 at 6:57
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.
– Amit Levy
Dec 3 at 6:57
add a comment |
1 Answer
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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 at 11:33
I like Serena
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1 Answer
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1 Answer
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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 at 11:33
I like Serena
3,6921718
add a comment |
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 at 11:33
I like Serena
3,6921718
add a comment |
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 at 11:33
I like Serena
3,6921718
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 at 11:33
I like Serena
3,6921718
answered Dec 2 at 11:33
I like Serena
3,6921718
answered Dec 2 at 11:33
I like Serena
3,6921718
answered Dec 2 at 11:33
I like Serena
3,6921718
3,6921718
add a comment |
add a comment |
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Have you calculated the answer for small values of $n$?
– Gerry Myerson
Dec 2 at 11:33
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.
– Amit Levy
Dec 3 at 6:57