How many different operations can be defined in a finite groupoid with a given property?
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Set $B=left{ 1, 2, ... 18 right}$ is given. How many different operations $*$ can be defined so that $(B,*)$ is a groupoid with a property that $|left{i|i*(19-i) neq i ∧ i*(19-i) neq (19-i)right}| = k$, where $k in B∪left{ 0 right}$.
What exactly is $k$ here? The solution is given as:
$binom{18}{k}(18-2)^k2^{18-k}18^{18^2-18}$.
I understand that the part within the parentheses say that eg. $18*1 neq 1 ∧ neq 18$ which explains the $18^{18^2-18}$
as a diagonal which follows these rules does not have 18 possibilities anymore. But what does this $k$ mean? And how does it change the result?
combinatorics elementary-set-theory magma
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
asked Jun 20 at 20:02
edward_d
1601211
add a comment |
up vote
4
down vote
favorite
Set $B=left{ 1, 2, ... 18 right}$ is given. How many different operations $*$ can be defined so that $(B,*)$ is a groupoid with a property that $|left{i|i*(19-i) neq i ∧ i*(19-i) neq (19-i)right}| = k$, where $k in B∪left{ 0 right}$.
What exactly is $k$ here? The solution is given as:
$binom{18}{k}(18-2)^k2^{18-k}18^{18^2-18}$.
I understand that the part within the parentheses say that eg. $18*1 neq 1 ∧ neq 18$ which explains the $18^{18^2-18}$
as a diagonal which follows these rules does not have 18 possibilities anymore. But what does this $k$ mean? And how does it change the result?
combinatorics elementary-set-theory magma
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
asked Jun 20 at 20:02
edward_d
1601211
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Set $B=left{ 1, 2, ... 18 right}$ is given. How many different operations $*$ can be defined so that $(B,*)$ is a groupoid with a property that $|left{i|i*(19-i) neq i ∧ i*(19-i) neq (19-i)right}| = k$, where $k in B∪left{ 0 right}$.
What exactly is $k$ here? The solution is given as:
$binom{18}{k}(18-2)^k2^{18-k}18^{18^2-18}$.
I understand that the part within the parentheses say that eg. $18*1 neq 1 ∧ neq 18$ which explains the $18^{18^2-18}$
as a diagonal which follows these rules does not have 18 possibilities anymore. But what does this $k$ mean? And how does it change the result?
combinatorics elementary-set-theory magma
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
asked Jun 20 at 20:02
edward_d
1601211
Set $B=left{ 1, 2, ... 18 right}$ is given. How many different operations $*$ can be defined so that $(B,*)$ is a groupoid with a property that $|left{i|i*(19-i) neq i ∧ i*(19-i) neq (19-i)right}| = k$, where $k in B∪left{ 0 right}$.
What exactly is $k$ here? The solution is given as:
$binom{18}{k}(18-2)^k2^{18-k}18^{18^2-18}$.
I understand that the part within the parentheses say that eg. $18*1 neq 1 ∧ neq 18$ which explains the $18^{18^2-18}$
as a diagonal which follows these rules does not have 18 possibilities anymore. But what does this $k$ mean? And how does it change the result?
combinatorics elementary-set-theory magma
combinatorics elementary-set-theory magma
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
asked Jun 20 at 20:02
edward_d
1601211
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
asked Jun 20 at 20:02
edward_d
1601211
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
edited Nov 21 at 14:39
Arnaud D.
15.6k52343
15.6k52343
asked Jun 20 at 20:02
edward_d
1601211
asked Jun 20 at 20:02
edward_d
1601211
asked Jun 20 at 20:02
edward_d
1601211
1601211
add a comment |
add a comment |
1 Answer
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$k$ is the cardinality of the set of elements $i$ such that $i*(19-i)notin{i,19-i}$; that is, it is the number of such elements.
Without constraints, there would be $18^{18^2}$ different operations, as we can independently assign one of $18$ results of the operation to each of the $18^2$ pairs of operands. Given the constraint on the cardinality of the set, we can choose the $k$ elements in the set in $binom{18}k$ ways, assign one of $18-2$ admissible results to their products of the form $i*(19-i)$, yielding a factor $(18-2)^k$, and for the remaining $18-k$ elements we have to assign one of the two inadmissible results $i$ and $19-i$ to their products of the form $i*(19-i)$, yielding a factor $2^{18-k}$. That leaves $18^2-18$ products to be freely chosen among the $18$ possibilities, yielding a factor $18^{18^2-18}$.
answered Jun 20 at 20:42
joriki
170k10183342
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
up vote
1
down vote
accepted
$k$ is the cardinality of the set of elements $i$ such that $i*(19-i)notin{i,19-i}$; that is, it is the number of such elements.
Without constraints, there would be $18^{18^2}$ different operations, as we can independently assign one of $18$ results of the operation to each of the $18^2$ pairs of operands. Given the constraint on the cardinality of the set, we can choose the $k$ elements in the set in $binom{18}k$ ways, assign one of $18-2$ admissible results to their products of the form $i*(19-i)$, yielding a factor $(18-2)^k$, and for the remaining $18-k$ elements we have to assign one of the two inadmissible results $i$ and $19-i$ to their products of the form $i*(19-i)$, yielding a factor $2^{18-k}$. That leaves $18^2-18$ products to be freely chosen among the $18$ possibilities, yielding a factor $18^{18^2-18}$.
answered Jun 20 at 20:42
joriki
170k10183342
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
add a comment |
up vote
1
down vote
accepted
$k$ is the cardinality of the set of elements $i$ such that $i*(19-i)notin{i,19-i}$; that is, it is the number of such elements.
Without constraints, there would be $18^{18^2}$ different operations, as we can independently assign one of $18$ results of the operation to each of the $18^2$ pairs of operands. Given the constraint on the cardinality of the set, we can choose the $k$ elements in the set in $binom{18}k$ ways, assign one of $18-2$ admissible results to their products of the form $i*(19-i)$, yielding a factor $(18-2)^k$, and for the remaining $18-k$ elements we have to assign one of the two inadmissible results $i$ and $19-i$ to their products of the form $i*(19-i)$, yielding a factor $2^{18-k}$. That leaves $18^2-18$ products to be freely chosen among the $18$ possibilities, yielding a factor $18^{18^2-18}$.
answered Jun 20 at 20:42
joriki
170k10183342
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$k$ is the cardinality of the set of elements $i$ such that $i*(19-i)notin{i,19-i}$; that is, it is the number of such elements.
Without constraints, there would be $18^{18^2}$ different operations, as we can independently assign one of $18$ results of the operation to each of the $18^2$ pairs of operands. Given the constraint on the cardinality of the set, we can choose the $k$ elements in the set in $binom{18}k$ ways, assign one of $18-2$ admissible results to their products of the form $i*(19-i)$, yielding a factor $(18-2)^k$, and for the remaining $18-k$ elements we have to assign one of the two inadmissible results $i$ and $19-i$ to their products of the form $i*(19-i)$, yielding a factor $2^{18-k}$. That leaves $18^2-18$ products to be freely chosen among the $18$ possibilities, yielding a factor $18^{18^2-18}$.
answered Jun 20 at 20:42
joriki
170k10183342
$k$ is the cardinality of the set of elements $i$ such that $i*(19-i)notin{i,19-i}$; that is, it is the number of such elements.
Without constraints, there would be $18^{18^2}$ different operations, as we can independently assign one of $18$ results of the operation to each of the $18^2$ pairs of operands. Given the constraint on the cardinality of the set, we can choose the $k$ elements in the set in $binom{18}k$ ways, assign one of $18-2$ admissible results to their products of the form $i*(19-i)$, yielding a factor $(18-2)^k$, and for the remaining $18-k$ elements we have to assign one of the two inadmissible results $i$ and $19-i$ to their products of the form $i*(19-i)$, yielding a factor $2^{18-k}$. That leaves $18^2-18$ products to be freely chosen among the $18$ possibilities, yielding a factor $18^{18^2-18}$.
answered Jun 20 at 20:42
joriki
170k10183342
edited Jun 20 at 21:28
answered Jun 20 at 20:42
joriki
170k10183342
answered Jun 20 at 20:42
joriki
170k10183342
answered Jun 20 at 20:42
joriki
170k10183342
170k10183342
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
add a comment |
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
Did you mean 18 over k ways to choose k and not 18 over 2?
– edward_d
Jun 20 at 20:52
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
@edward_d: I did, thanks; I fixed it.
– joriki
Jun 20 at 21:28
add a comment |
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