Mixing
how many partitions and equivalence relations of A = a,b,c exists for example or A a,b,c,d,e?
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I have these two questions. Is there a formel or anything like that to find how many partitions and equivalence relations there is for A? I know the answer but i don´t know how to calculate it by my self.
I would appreciate if anyone could explain how it works.
equivalence-relations set-partition
asked Jul 31 at 13:36
thpthp
1
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up vote
0
down vote
favorite
I have these two questions. Is there a formel or anything like that to find how many partitions and equivalence relations there is for A? I know the answer but i don´t know how to calculate it by my self.
I would appreciate if anyone could explain how it works.
equivalence-relations set-partition
asked Jul 31 at 13:36
thpthp
1
1
en.wikipedia.org/wiki/Bell_number
– saulspatz
Jul 31 at 13:40
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 13:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have these two questions. Is there a formel or anything like that to find how many partitions and equivalence relations there is for A? I know the answer but i don´t know how to calculate it by my self.
I would appreciate if anyone could explain how it works.
equivalence-relations set-partition
asked Jul 31 at 13:36
thpthp
1
I have these two questions. Is there a formel or anything like that to find how many partitions and equivalence relations there is for A? I know the answer but i don´t know how to calculate it by my self.
I would appreciate if anyone could explain how it works.
equivalence-relations set-partition
asked Jul 31 at 13:36
thpthp
1
asked Jul 31 at 13:36
thpthp
1
asked Jul 31 at 13:36
thpthp
1
asked Jul 31 at 13:36
thpthp
1
1
1
en.wikipedia.org/wiki/Bell_number
– saulspatz
Jul 31 at 13:40
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 13:42
add a comment |Â
1
en.wikipedia.org/wiki/Bell_number
– saulspatz
Jul 31 at 13:40
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 13:42
1
1
en.wikipedia.org/wiki/Bell_number
– saulspatz
Jul 31 at 13:40
en.wikipedia.org/wiki/Bell_number
– saulspatz
Jul 31 at 13:40
1
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 13:42
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 13:42
add a comment |Â
1 Answer
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You have this recurrence relation
To prove it, for a set of $n+1$ elements, single out one element $a$. Then there are $binomnn-k=binomnk$ ways to choose $n-k$ other elements among the other $n$ elements to be in the class of $a$, when that class ought to have $n-k$ elements. For each of those there are $B_k$ ways to partition the remaining $k$ elements. Summing for all the cases for the size $n-k=0,1,...,n$ of the class of $a$, you get the value of $B_n+1$.
answered Jul 31 at 13:50
plasticConnection
362
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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
up vote
0
down vote
You have this recurrence relation
To prove it, for a set of $n+1$ elements, single out one element $a$. Then there are $binomnn-k=binomnk$ ways to choose $n-k$ other elements among the other $n$ elements to be in the class of $a$, when that class ought to have $n-k$ elements. For each of those there are $B_k$ ways to partition the remaining $k$ elements. Summing for all the cases for the size $n-k=0,1,...,n$ of the class of $a$, you get the value of $B_n+1$.
answered Jul 31 at 13:50
plasticConnection
362
add a comment |Â
up vote
0
down vote
You have this recurrence relation
To prove it, for a set of $n+1$ elements, single out one element $a$. Then there are $binomnn-k=binomnk$ ways to choose $n-k$ other elements among the other $n$ elements to be in the class of $a$, when that class ought to have $n-k$ elements. For each of those there are $B_k$ ways to partition the remaining $k$ elements. Summing for all the cases for the size $n-k=0,1,...,n$ of the class of $a$, you get the value of $B_n+1$.
answered Jul 31 at 13:50
plasticConnection
362
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have this recurrence relation
To prove it, for a set of $n+1$ elements, single out one element $a$. Then there are $binomnn-k=binomnk$ ways to choose $n-k$ other elements among the other $n$ elements to be in the class of $a$, when that class ought to have $n-k$ elements. For each of those there are $B_k$ ways to partition the remaining $k$ elements. Summing for all the cases for the size $n-k=0,1,...,n$ of the class of $a$, you get the value of $B_n+1$.
answered Jul 31 at 13:50
plasticConnection
362
You have this recurrence relation
To prove it, for a set of $n+1$ elements, single out one element $a$. Then there are $binomnn-k=binomnk$ ways to choose $n-k$ other elements among the other $n$ elements to be in the class of $a$, when that class ought to have $n-k$ elements. For each of those there are $B_k$ ways to partition the remaining $k$ elements. Summing for all the cases for the size $n-k=0,1,...,n$ of the class of $a$, you get the value of $B_n+1$.
answered Jul 31 at 13:50
plasticConnection
362
answered Jul 31 at 13:50
plasticConnection
362
answered Jul 31 at 13:50
plasticConnection
362
answered Jul 31 at 13:50
plasticConnection
362
362
add a comment |Â
add a comment |Â
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1
en.wikipedia.org/wiki/Bell_number
– saulspatz
Jul 31 at 13:40
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 13:42