Using subfactorial in algebra
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Can someone explain the real use of subfactorial? I know that factorial and subfactorial are related to each other, but they function differently.
combinatorics factorial
edited Sep 6 at 9:37
David G. Stork
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asked Sep 6 at 8:13
MMJM
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up vote
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Can someone explain the real use of subfactorial? I know that factorial and subfactorial are related to each other, but they function differently.
combinatorics factorial
edited Sep 6 at 9:37
David G. Stork
8,15621232
asked Sep 6 at 8:13
MMJM
1469
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can someone explain the real use of subfactorial? I know that factorial and subfactorial are related to each other, but they function differently.
combinatorics factorial
edited Sep 6 at 9:37
David G. Stork
8,15621232
asked Sep 6 at 8:13
MMJM
1469
Can someone explain the real use of subfactorial? I know that factorial and subfactorial are related to each other, but they function differently.
combinatorics factorial
combinatorics factorial
edited Sep 6 at 9:37
David G. Stork
8,15621232
asked Sep 6 at 8:13
MMJM
1469
edited Sep 6 at 9:37
David G. Stork
8,15621232
asked Sep 6 at 8:13
MMJM
1469
edited Sep 6 at 9:37
David G. Stork
8,15621232
edited Sep 6 at 9:37
David G. Stork
8,15621232
edited Sep 6 at 9:37
David G. Stork
8,15621232
8,15621232
asked Sep 6 at 8:13
MMJM
1469
asked Sep 6 at 8:13
MMJM
1469
asked Sep 6 at 8:13
MMJM
1469
1469
add a comment |Â
add a comment |Â
1 Answer
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The subfactorial is useful in calculating the number of derangements, i.e., the number of permutation of $n$ objects in which none of them remain in their original position. For instance, one derangement of $ABCDEF$ is $BADCFE$. (The permutation $BADECF$ is not a derangement of $ABCDEF$ because $F$ remains in the last position.)
The number of derangements of $n$ objects is:
$$!n = (n-1)(!(n-1) + !(n-2)) ,$$
where $!n$ is read "$n$ subfactorial."
For $n = 1, ldots , 10$ the values are $0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961 $.
For the $n=6$ case of $ABCDEF$, we find $!n = 265$.
answered Sep 6 at 9:10
David G. Stork
8,15621232
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
The subfactorial is useful in calculating the number of derangements, i.e., the number of permutation of $n$ objects in which none of them remain in their original position. For instance, one derangement of $ABCDEF$ is $BADCFE$. (The permutation $BADECF$ is not a derangement of $ABCDEF$ because $F$ remains in the last position.)
The number of derangements of $n$ objects is:
$$!n = (n-1)(!(n-1) + !(n-2)) ,$$
where $!n$ is read "$n$ subfactorial."
For $n = 1, ldots , 10$ the values are $0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961 $.
For the $n=6$ case of $ABCDEF$, we find $!n = 265$.
answered Sep 6 at 9:10
David G. Stork
8,15621232
add a comment |Â
up vote
1
down vote
accepted
The subfactorial is useful in calculating the number of derangements, i.e., the number of permutation of $n$ objects in which none of them remain in their original position. For instance, one derangement of $ABCDEF$ is $BADCFE$. (The permutation $BADECF$ is not a derangement of $ABCDEF$ because $F$ remains in the last position.)
The number of derangements of $n$ objects is:
$$!n = (n-1)(!(n-1) + !(n-2)) ,$$
where $!n$ is read "$n$ subfactorial."
For $n = 1, ldots , 10$ the values are $0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961 $.
For the $n=6$ case of $ABCDEF$, we find $!n = 265$.
answered Sep 6 at 9:10
David G. Stork
8,15621232
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The subfactorial is useful in calculating the number of derangements, i.e., the number of permutation of $n$ objects in which none of them remain in their original position. For instance, one derangement of $ABCDEF$ is $BADCFE$. (The permutation $BADECF$ is not a derangement of $ABCDEF$ because $F$ remains in the last position.)
The number of derangements of $n$ objects is:
$$!n = (n-1)(!(n-1) + !(n-2)) ,$$
where $!n$ is read "$n$ subfactorial."
For $n = 1, ldots , 10$ the values are $0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961 $.
For the $n=6$ case of $ABCDEF$, we find $!n = 265$.
answered Sep 6 at 9:10
David G. Stork
8,15621232
The subfactorial is useful in calculating the number of derangements, i.e., the number of permutation of $n$ objects in which none of them remain in their original position. For instance, one derangement of $ABCDEF$ is $BADCFE$. (The permutation $BADECF$ is not a derangement of $ABCDEF$ because $F$ remains in the last position.)
The number of derangements of $n$ objects is:
$$!n = (n-1)(!(n-1) + !(n-2)) ,$$
where $!n$ is read "$n$ subfactorial."
For $n = 1, ldots , 10$ the values are $0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961 $.
For the $n=6$ case of $ABCDEF$, we find $!n = 265$.
answered Sep 6 at 9:10
David G. Stork
8,15621232
edited Sep 6 at 9:17
answered Sep 6 at 9:10
David G. Stork
8,15621232
answered Sep 6 at 9:10
David G. Stork
8,15621232
answered Sep 6 at 9:10
David G. Stork
8,15621232
8,15621232
add a comment |Â
add a comment |Â
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