Mixing
Permutations permuting all elements.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Given $S_n$ acting over $T=1,2,dots,n$ in the usual way, can I count the number of elements in $S_n-cup_i=1^n G_i$ (where $G_i$ is the stabilizer of the element $iin T$) and get an explicit form for them (I mean obtain certain elements that generates this set)?
For example: for $n=3$, the elements of $S_3$ are $(1,2), (1,3), (2,3), (1,2,3), (1,3,2), ()$, so the only elements that are in this set are $(1,2,3)$ and $(1,3,2)$.
group-theory finite-groups permutations
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
add a comment |Â
up vote
1
down vote
favorite
Given $S_n$ acting over $T=1,2,dots,n$ in the usual way, can I count the number of elements in $S_n-cup_i=1^n G_i$ (where $G_i$ is the stabilizer of the element $iin T$) and get an explicit form for them (I mean obtain certain elements that generates this set)?
For example: for $n=3$, the elements of $S_3$ are $(1,2), (1,3), (2,3), (1,2,3), (1,3,2), ()$, so the only elements that are in this set are $(1,2,3)$ and $(1,3,2)$.
group-theory finite-groups permutations
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $S_n$ acting over $T=1,2,dots,n$ in the usual way, can I count the number of elements in $S_n-cup_i=1^n G_i$ (where $G_i$ is the stabilizer of the element $iin T$) and get an explicit form for them (I mean obtain certain elements that generates this set)?
For example: for $n=3$, the elements of $S_3$ are $(1,2), (1,3), (2,3), (1,2,3), (1,3,2), ()$, so the only elements that are in this set are $(1,2,3)$ and $(1,3,2)$.
group-theory finite-groups permutations
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
Given $S_n$ acting over $T=1,2,dots,n$ in the usual way, can I count the number of elements in $S_n-cup_i=1^n G_i$ (where $G_i$ is the stabilizer of the element $iin T$) and get an explicit form for them (I mean obtain certain elements that generates this set)?
For example: for $n=3$, the elements of $S_3$ are $(1,2), (1,3), (2,3), (1,2,3), (1,3,2), ()$, so the only elements that are in this set are $(1,2,3)$ and $(1,3,2)$.
group-theory finite-groups permutations
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
asked Jul 22 at 4:49
MonsieurGalois
3,3961233
3,3961233
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Yes, these are known as derangements. Using the principle of inclusion-exclusion, you can show that the number of them is $$n!sum_k =0^n frac(-1)^kk!,.$$
answered Jul 22 at 5:02
Marcus M
8,1731847
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, these are known as derangements. Using the principle of inclusion-exclusion, you can show that the number of them is $$n!sum_k =0^n frac(-1)^kk!,.$$
answered Jul 22 at 5:02
Marcus M
8,1731847
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
add a comment |Â
up vote
1
down vote
accepted
Yes, these are known as derangements. Using the principle of inclusion-exclusion, you can show that the number of them is $$n!sum_k =0^n frac(-1)^kk!,.$$
answered Jul 22 at 5:02
Marcus M
8,1731847
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, these are known as derangements. Using the principle of inclusion-exclusion, you can show that the number of them is $$n!sum_k =0^n frac(-1)^kk!,.$$
answered Jul 22 at 5:02
Marcus M
8,1731847
Yes, these are known as derangements. Using the principle of inclusion-exclusion, you can show that the number of them is $$n!sum_k =0^n frac(-1)^kk!,.$$
answered Jul 22 at 5:02
Marcus M
8,1731847
answered Jul 22 at 5:02
Marcus M
8,1731847
answered Jul 22 at 5:02
Marcus M
8,1731847
answered Jul 22 at 5:02
Marcus M
8,1731847
8,1731847
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
add a comment |Â
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
You might add that the proportion of derangements converses very rapidly to $e^-1$. I remember encountering this when very young and wondering about the probability of getting a snap when playing snap with two normal decks of $52$ cards. It's a situation where the number $e$ arises in "real life".
– Derek Holt
Jul 22 at 10:11
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859101%2fpermutations-permuting-all-elements%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password