Let $C_n $ be the number of ways for $n$ people to form several nonempty lines and then to arrange these...
$begingroup$
Let $C_n$ be the number of ways for $n$ people to form several nonempty lines, and then to arrange these lines in a circular order. Find a closed formula for $C_n$.
I was thinking to use either exponential or compositional formula, but I am very unsure about this problem. Please help me on this.
combinatorics
edited Dec 17 '18 at 6:31
dmtri
1,4542521
asked Dec 17 '18 at 5:26
MojicaMojica
202
$endgroup$
add a comment |
$begingroup$
Let $C_n$ be the number of ways for $n$ people to form several nonempty lines, and then to arrange these lines in a circular order. Find a closed formula for $C_n$.
I was thinking to use either exponential or compositional formula, but I am very unsure about this problem. Please help me on this.
combinatorics
edited Dec 17 '18 at 6:31
dmtri
1,4542521
asked Dec 17 '18 at 5:26
MojicaMojica
202
$endgroup$
add a comment |
$begingroup$
Let $C_n$ be the number of ways for $n$ people to form several nonempty lines, and then to arrange these lines in a circular order. Find a closed formula for $C_n$.
I was thinking to use either exponential or compositional formula, but I am very unsure about this problem. Please help me on this.
combinatorics
edited Dec 17 '18 at 6:31
dmtri
1,4542521
asked Dec 17 '18 at 5:26
MojicaMojica
202
$endgroup$
Let $C_n$ be the number of ways for $n$ people to form several nonempty lines, and then to arrange these lines in a circular order. Find a closed formula for $C_n$.
I was thinking to use either exponential or compositional formula, but I am very unsure about this problem. Please help me on this.
combinatorics
combinatorics
edited Dec 17 '18 at 6:31
dmtri
1,4542521
asked Dec 17 '18 at 5:26
MojicaMojica
202
edited Dec 17 '18 at 6:31
dmtri
1,4542521
asked Dec 17 '18 at 5:26
MojicaMojica
202
edited Dec 17 '18 at 6:31
dmtri
1,4542521
edited Dec 17 '18 at 6:31
dmtri
1,4542521
edited Dec 17 '18 at 6:31
dmtri
1,4542521
1,4542521
asked Dec 17 '18 at 5:26
MojicaMojica
202
asked Dec 17 '18 at 5:26
MojicaMojica
202
asked Dec 17 '18 at 5:26
MojicaMojica
202
202
add a comment |
add a comment |
1 Answer
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$begingroup$
The exponential generating function for a nonempty line is $x/(1-x)$. The egf for forming a linear order of $k$ non-empty lines is therefore $(x/(1-x))^k$; we must then divide by $k$ to make this a cyclic order. Summing over $k$, and recalling $sum_{kge1}y^k/k=-log(1-y)$, we get the egf for $C_n$ is $-log(1-x/(1-x))$. I leave it to you to extract the formula for $C_n$.
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
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add a comment |
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$begingroup$
The exponential generating function for a nonempty line is $x/(1-x)$. The egf for forming a linear order of $k$ non-empty lines is therefore $(x/(1-x))^k$; we must then divide by $k$ to make this a cyclic order. Summing over $k$, and recalling $sum_{kge1}y^k/k=-log(1-y)$, we get the egf for $C_n$ is $-log(1-x/(1-x))$. I leave it to you to extract the formula for $C_n$.
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
$endgroup$
add a comment |
$begingroup$
The exponential generating function for a nonempty line is $x/(1-x)$. The egf for forming a linear order of $k$ non-empty lines is therefore $(x/(1-x))^k$; we must then divide by $k$ to make this a cyclic order. Summing over $k$, and recalling $sum_{kge1}y^k/k=-log(1-y)$, we get the egf for $C_n$ is $-log(1-x/(1-x))$. I leave it to you to extract the formula for $C_n$.
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
$endgroup$
add a comment |
$begingroup$
The exponential generating function for a nonempty line is $x/(1-x)$. The egf for forming a linear order of $k$ non-empty lines is therefore $(x/(1-x))^k$; we must then divide by $k$ to make this a cyclic order. Summing over $k$, and recalling $sum_{kge1}y^k/k=-log(1-y)$, we get the egf for $C_n$ is $-log(1-x/(1-x))$. I leave it to you to extract the formula for $C_n$.
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
$endgroup$
The exponential generating function for a nonempty line is $x/(1-x)$. The egf for forming a linear order of $k$ non-empty lines is therefore $(x/(1-x))^k$; we must then divide by $k$ to make this a cyclic order. Summing over $k$, and recalling $sum_{kge1}y^k/k=-log(1-y)$, we get the egf for $C_n$ is $-log(1-x/(1-x))$. I leave it to you to extract the formula for $C_n$.
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
answered Dec 17 '18 at 16:02
Mike EarnestMike Earnest
21.3k11951
21.3k11951
add a comment |
add a comment |
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