Number of permutations that map every part of a particular partition away from itself
Let $C_1={1,2,dots,c_1}, space C_2={c_1+1, dots,c_2}, dots space C_k={c_{k-1}+1,dots, n}$ be a partition of the set ${1,2,dots n}$.
I want to calculate the number of permutations $sigma in S_n$ that don't take any element in any part of the partition to that same part; that is, the number
$$a_C = #{sigmain S_n space | space forall k : forall j in C_k : sigma(j) notin C_k}$$
My thoughts and "conjectures" on the problem:
A special case when $|C_j|=d$ is constant and $n=kd$.
A further special case of this is $d=4, k=13$ (and $n=52$). This corresponds to a card game where the deck is gone through one card at a time and at the same time counted $1,2,dots, 13,1,2,dots, 13,1,2,dots, 13,1,2,dots, 13 $. If at any point the number said out loud matches the value of the card laid down (suit doesn't matter), the game is lost. Otherwise, if the end is reached without matches, the game is won. (Trying to calculate the winning probability of this game is how I came up with the general problem.)
The case $d=1$ gives the permutations without fixed points. For these the fraction $frac{a_n}{n!} to frac{1}{e}$ as $nto infty$. I suspect that for other values of $d$ the limit is $frac{1}{e^d}$ but I can't find a proof for this. Still, simulation seems to support the claim and there's the heuristic argument that if the events $sigma(j) notin C_k$ were independent, the probability of "$sigma$ to be good" would be $(1-frac{d}{n})^n to e^{-d}$.
For some partitions $a_C=0$: if one part has more than $n/2$ elements, some element must stay in that part after permuting.
probability permutations
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
add a comment |
Let $C_1={1,2,dots,c_1}, space C_2={c_1+1, dots,c_2}, dots space C_k={c_{k-1}+1,dots, n}$ be a partition of the set ${1,2,dots n}$.
I want to calculate the number of permutations $sigma in S_n$ that don't take any element in any part of the partition to that same part; that is, the number
$$a_C = #{sigmain S_n space | space forall k : forall j in C_k : sigma(j) notin C_k}$$
My thoughts and "conjectures" on the problem:
A special case when $|C_j|=d$ is constant and $n=kd$.
A further special case of this is $d=4, k=13$ (and $n=52$). This corresponds to a card game where the deck is gone through one card at a time and at the same time counted $1,2,dots, 13,1,2,dots, 13,1,2,dots, 13,1,2,dots, 13 $. If at any point the number said out loud matches the value of the card laid down (suit doesn't matter), the game is lost. Otherwise, if the end is reached without matches, the game is won. (Trying to calculate the winning probability of this game is how I came up with the general problem.)
The case $d=1$ gives the permutations without fixed points. For these the fraction $frac{a_n}{n!} to frac{1}{e}$ as $nto infty$. I suspect that for other values of $d$ the limit is $frac{1}{e^d}$ but I can't find a proof for this. Still, simulation seems to support the claim and there's the heuristic argument that if the events $sigma(j) notin C_k$ were independent, the probability of "$sigma$ to be good" would be $(1-frac{d}{n})^n to e^{-d}$.
For some partitions $a_C=0$: if one part has more than $n/2$ elements, some element must stay in that part after permuting.
probability permutations
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
add a comment |
Let $C_1={1,2,dots,c_1}, space C_2={c_1+1, dots,c_2}, dots space C_k={c_{k-1}+1,dots, n}$ be a partition of the set ${1,2,dots n}$.
I want to calculate the number of permutations $sigma in S_n$ that don't take any element in any part of the partition to that same part; that is, the number
$$a_C = #{sigmain S_n space | space forall k : forall j in C_k : sigma(j) notin C_k}$$
My thoughts and "conjectures" on the problem:
A special case when $|C_j|=d$ is constant and $n=kd$.
A further special case of this is $d=4, k=13$ (and $n=52$). This corresponds to a card game where the deck is gone through one card at a time and at the same time counted $1,2,dots, 13,1,2,dots, 13,1,2,dots, 13,1,2,dots, 13 $. If at any point the number said out loud matches the value of the card laid down (suit doesn't matter), the game is lost. Otherwise, if the end is reached without matches, the game is won. (Trying to calculate the winning probability of this game is how I came up with the general problem.)
The case $d=1$ gives the permutations without fixed points. For these the fraction $frac{a_n}{n!} to frac{1}{e}$ as $nto infty$. I suspect that for other values of $d$ the limit is $frac{1}{e^d}$ but I can't find a proof for this. Still, simulation seems to support the claim and there's the heuristic argument that if the events $sigma(j) notin C_k$ were independent, the probability of "$sigma$ to be good" would be $(1-frac{d}{n})^n to e^{-d}$.
For some partitions $a_C=0$: if one part has more than $n/2$ elements, some element must stay in that part after permuting.
probability permutations
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
Let $C_1={1,2,dots,c_1}, space C_2={c_1+1, dots,c_2}, dots space C_k={c_{k-1}+1,dots, n}$ be a partition of the set ${1,2,dots n}$.
I want to calculate the number of permutations $sigma in S_n$ that don't take any element in any part of the partition to that same part; that is, the number
$$a_C = #{sigmain S_n space | space forall k : forall j in C_k : sigma(j) notin C_k}$$
My thoughts and "conjectures" on the problem:
A special case when $|C_j|=d$ is constant and $n=kd$.
A further special case of this is $d=4, k=13$ (and $n=52$). This corresponds to a card game where the deck is gone through one card at a time and at the same time counted $1,2,dots, 13,1,2,dots, 13,1,2,dots, 13,1,2,dots, 13 $. If at any point the number said out loud matches the value of the card laid down (suit doesn't matter), the game is lost. Otherwise, if the end is reached without matches, the game is won. (Trying to calculate the winning probability of this game is how I came up with the general problem.)
The case $d=1$ gives the permutations without fixed points. For these the fraction $frac{a_n}{n!} to frac{1}{e}$ as $nto infty$. I suspect that for other values of $d$ the limit is $frac{1}{e^d}$ but I can't find a proof for this. Still, simulation seems to support the claim and there's the heuristic argument that if the events $sigma(j) notin C_k$ were independent, the probability of "$sigma$ to be good" would be $(1-frac{d}{n})^n to e^{-d}$.
For some partitions $a_C=0$: if one part has more than $n/2$ elements, some element must stay in that part after permuting.
probability permutations
probability permutations
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
asked Jun 13 '17 at 19:35
ploosu2ploosu2
4,5771023
4,5771023
add a comment |
add a comment |
1 Answer
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The answer is given by
$$a_C = int_0^infty prod_{j=1}^k m_j! L_{m_j}(x) e^{-x} dx$$
where $m_j = |C_j|$ and $L_t$ is $t$th Laguerre Polynomial (up to a sign).
See the wikipedia entry on derangement generalizations. Notice here we distinguish the elements inside each class $C_j$, on contrast to Wikipedia's case where words and letters are considered. Therefore we multiply by $m_j!$ here (also notice: this is precisely the constant factor in front of each Laguerre Polynomial, so this amounts to just dropping that factor).
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The answer is given by
$$a_C = int_0^infty prod_{j=1}^k m_j! L_{m_j}(x) e^{-x} dx$$
where $m_j = |C_j|$ and $L_t$ is $t$th Laguerre Polynomial (up to a sign).
See the wikipedia entry on derangement generalizations. Notice here we distinguish the elements inside each class $C_j$, on contrast to Wikipedia's case where words and letters are considered. Therefore we multiply by $m_j!$ here (also notice: this is precisely the constant factor in front of each Laguerre Polynomial, so this amounts to just dropping that factor).
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
add a comment |
The answer is given by
$$a_C = int_0^infty prod_{j=1}^k m_j! L_{m_j}(x) e^{-x} dx$$
where $m_j = |C_j|$ and $L_t$ is $t$th Laguerre Polynomial (up to a sign).
See the wikipedia entry on derangement generalizations. Notice here we distinguish the elements inside each class $C_j$, on contrast to Wikipedia's case where words and letters are considered. Therefore we multiply by $m_j!$ here (also notice: this is precisely the constant factor in front of each Laguerre Polynomial, so this amounts to just dropping that factor).
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
add a comment |
The answer is given by
$$a_C = int_0^infty prod_{j=1}^k m_j! L_{m_j}(x) e^{-x} dx$$
where $m_j = |C_j|$ and $L_t$ is $t$th Laguerre Polynomial (up to a sign).
See the wikipedia entry on derangement generalizations. Notice here we distinguish the elements inside each class $C_j$, on contrast to Wikipedia's case where words and letters are considered. Therefore we multiply by $m_j!$ here (also notice: this is precisely the constant factor in front of each Laguerre Polynomial, so this amounts to just dropping that factor).
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
The answer is given by
$$a_C = int_0^infty prod_{j=1}^k m_j! L_{m_j}(x) e^{-x} dx$$
where $m_j = |C_j|$ and $L_t$ is $t$th Laguerre Polynomial (up to a sign).
See the wikipedia entry on derangement generalizations. Notice here we distinguish the elements inside each class $C_j$, on contrast to Wikipedia's case where words and letters are considered. Therefore we multiply by $m_j!$ here (also notice: this is precisely the constant factor in front of each Laguerre Polynomial, so this amounts to just dropping that factor).
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
answered Dec 5 '18 at 10:20
ploosu2ploosu2
4,5771023
4,5771023
add a comment |
add a comment |
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