What's the formula for partitions where each group has a different size? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How does distinguishability of boxes change the number of ways to distribute n objects into separate boxesHow many different sets of groups can be formed so that nobody ever has the same group mates?Partitioning $n$ objects into $k$ boxesGroup forming when the group size is equalSample, randomly & uniformly the partitioning of $n$ objects into $K$ groupsArgue that $binomnn_1,n_2,…,n_r = binomn-1n_1-1,n_2,…,n_r + binomn-1n_1,n_2-1,…,n_r+…+binomn-1n_1,n_2,…,n_r-1 $C compositions of $N$ balls grouped in k types given first and/or last offset …How many ordered sequences of length $n$ can we make with $n_i$ identical objects of each type $i$?Probability that two particular items are grouped in a random partition with fixed sizesAre Complete Sequences Also Sets with Distinct Subset Sums? (Prime Numbers)Number of ways to divide a group of n people into groups of size m to m-1
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What's the formula for partitions where each group has a different size?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How does distinguishability of boxes change the number of ways to distribute n objects into separate boxesHow many different sets of groups can be formed so that nobody ever has the same group mates?Partitioning $n$ objects into $k$ boxesGroup forming when the group size is equalSample, randomly & uniformly the partitioning of $n$ objects into $K$ groupsArgue that $binomnn_1,n_2,…,n_r = binomn-1n_1-1,n_2,…,n_r + binomn-1n_1,n_2-1,…,n_r+…+binomn-1n_1,n_2,…,n_r-1 $C compositions of $N$ balls grouped in k types given first and/or last offset …How many ordered sequences of length $n$ can we make with $n_i$ identical objects of each type $i$?Probability that two particular items are grouped in a random partition with fixed sizesAre Complete Sequences Also Sets with Distinct Subset Sums? (Prime Numbers)Number of ways to divide a group of n people into groups of size m to m-1
$begingroup$
I am looking at this formula
My understanding is that the number of partitions is equal to that formula, but only when the order of the partitions matters. eg, That's right, if you can only have $n_1$ in the first group and $n_2$ in the second group and so on.
My understanding is that if the order of the groups does not matter, then it is the same formula except you add $r!$ to the denominator because you need to account for the different orderings that can achieve that result.
Is that right?
combinatorics
asked Apr 2 at 1:01
SebastianSebastian
1486
$endgroup$
add a comment |
$begingroup$
I am looking at this formula
My understanding is that the number of partitions is equal to that formula, but only when the order of the partitions matters. eg, That's right, if you can only have $n_1$ in the first group and $n_2$ in the second group and so on.
My understanding is that if the order of the groups does not matter, then it is the same formula except you add $r!$ to the denominator because you need to account for the different orderings that can achieve that result.
Is that right?
combinatorics
asked Apr 2 at 1:01
SebastianSebastian
1486
$endgroup$
$begingroup$
That's correct.
$endgroup$
– Jair Taylor
Apr 2 at 1:40
$begingroup$
Does the denominator need to add up to n in the first case?
$endgroup$
– Sebastian
Apr 2 at 2:13
$begingroup$
Yes, you need $sum_i=1^r n_i = n$ since the sizes of the blocks must add up to the size of the set
$endgroup$
– Jair Taylor
Apr 2 at 2:41
add a comment |
$begingroup$
I am looking at this formula
My understanding is that the number of partitions is equal to that formula, but only when the order of the partitions matters. eg, That's right, if you can only have $n_1$ in the first group and $n_2$ in the second group and so on.
My understanding is that if the order of the groups does not matter, then it is the same formula except you add $r!$ to the denominator because you need to account for the different orderings that can achieve that result.
Is that right?
combinatorics
asked Apr 2 at 1:01
SebastianSebastian
1486
$endgroup$
I am looking at this formula
My understanding is that the number of partitions is equal to that formula, but only when the order of the partitions matters. eg, That's right, if you can only have $n_1$ in the first group and $n_2$ in the second group and so on.
My understanding is that if the order of the groups does not matter, then it is the same formula except you add $r!$ to the denominator because you need to account for the different orderings that can achieve that result.
Is that right?
combinatorics
combinatorics
asked Apr 2 at 1:01
SebastianSebastian
1486
asked Apr 2 at 1:01
SebastianSebastian
1486
asked Apr 2 at 1:01
SebastianSebastian
1486
asked Apr 2 at 1:01
SebastianSebastian
1486
asked Apr 2 at 1:01
SebastianSebastian
1486
1486
$begingroup$
That's correct.
$endgroup$
– Jair Taylor
Apr 2 at 1:40
$begingroup$
Does the denominator need to add up to n in the first case?
$endgroup$
– Sebastian
Apr 2 at 2:13
$begingroup$
Yes, you need $sum_i=1^r n_i = n$ since the sizes of the blocks must add up to the size of the set
$endgroup$
– Jair Taylor
Apr 2 at 2:41
add a comment |
$begingroup$
That's correct.
$endgroup$
– Jair Taylor
Apr 2 at 1:40
$begingroup$
Does the denominator need to add up to n in the first case?
$endgroup$
– Sebastian
Apr 2 at 2:13
$begingroup$
Yes, you need $sum_i=1^r n_i = n$ since the sizes of the blocks must add up to the size of the set
$endgroup$
– Jair Taylor
Apr 2 at 2:41
$begingroup$
That's correct.
$endgroup$
– Jair Taylor
Apr 2 at 1:40
$begingroup$
That's correct.
$endgroup$
– Jair Taylor
Apr 2 at 1:40
$begingroup$
Does the denominator need to add up to n in the first case?
$endgroup$
– Sebastian
Apr 2 at 2:13
$begingroup$
Does the denominator need to add up to n in the first case?
$endgroup$
– Sebastian
Apr 2 at 2:13
$begingroup$
Yes, you need $sum_i=1^r n_i = n$ since the sizes of the blocks must add up to the size of the set
$endgroup$
– Jair Taylor
Apr 2 at 2:41
$begingroup$
Yes, you need $sum_i=1^r n_i = n$ since the sizes of the blocks must add up to the size of the set
$endgroup$
– Jair Taylor
Apr 2 at 2:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not correct. If all the groups are different sizes, then the answer is $fracn!n_1!n_2!dots n_r!$ regardless whether you care about the order of the groups. It cannot be the case that you divide by $r!$, since for example when $n=3,n_1=1,n_2=2$, then $frac3!1!2!=3$ is not divisible by $2!$.
If all groups are the same size, the you do need to add $r!$ to the denominator to discount for order, so the formula is $fracn!r!(n/r)!^r$.
In general, suppose the numbers $n_i$ are partitioned into several groups, where the $k^th$ group has size $lambda_k$, and the $n_i=n_j$ if and only if $n_i$ and $n_j$ are in the same group. Then the number of ways to partition the objects into groups of size $n_1,dots,n_r$ without respect to order is
$$
fracn!left(prod_i=1^r n_i!right)left(prod_k=1^m lambda_j!right).
$$
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
$endgroup$
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
This is not correct. If all the groups are different sizes, then the answer is $fracn!n_1!n_2!dots n_r!$ regardless whether you care about the order of the groups. It cannot be the case that you divide by $r!$, since for example when $n=3,n_1=1,n_2=2$, then $frac3!1!2!=3$ is not divisible by $2!$.
If all groups are the same size, the you do need to add $r!$ to the denominator to discount for order, so the formula is $fracn!r!(n/r)!^r$.
In general, suppose the numbers $n_i$ are partitioned into several groups, where the $k^th$ group has size $lambda_k$, and the $n_i=n_j$ if and only if $n_i$ and $n_j$ are in the same group. Then the number of ways to partition the objects into groups of size $n_1,dots,n_r$ without respect to order is
$$
fracn!left(prod_i=1^r n_i!right)left(prod_k=1^m lambda_j!right).
$$
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
$endgroup$
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
add a comment |
$begingroup$
This is not correct. If all the groups are different sizes, then the answer is $fracn!n_1!n_2!dots n_r!$ regardless whether you care about the order of the groups. It cannot be the case that you divide by $r!$, since for example when $n=3,n_1=1,n_2=2$, then $frac3!1!2!=3$ is not divisible by $2!$.
If all groups are the same size, the you do need to add $r!$ to the denominator to discount for order, so the formula is $fracn!r!(n/r)!^r$.
In general, suppose the numbers $n_i$ are partitioned into several groups, where the $k^th$ group has size $lambda_k$, and the $n_i=n_j$ if and only if $n_i$ and $n_j$ are in the same group. Then the number of ways to partition the objects into groups of size $n_1,dots,n_r$ without respect to order is
$$
fracn!left(prod_i=1^r n_i!right)left(prod_k=1^m lambda_j!right).
$$
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
$endgroup$
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
add a comment |
$begingroup$
This is not correct. If all the groups are different sizes, then the answer is $fracn!n_1!n_2!dots n_r!$ regardless whether you care about the order of the groups. It cannot be the case that you divide by $r!$, since for example when $n=3,n_1=1,n_2=2$, then $frac3!1!2!=3$ is not divisible by $2!$.
If all groups are the same size, the you do need to add $r!$ to the denominator to discount for order, so the formula is $fracn!r!(n/r)!^r$.
In general, suppose the numbers $n_i$ are partitioned into several groups, where the $k^th$ group has size $lambda_k$, and the $n_i=n_j$ if and only if $n_i$ and $n_j$ are in the same group. Then the number of ways to partition the objects into groups of size $n_1,dots,n_r$ without respect to order is
$$
fracn!left(prod_i=1^r n_i!right)left(prod_k=1^m lambda_j!right).
$$
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
$endgroup$
This is not correct. If all the groups are different sizes, then the answer is $fracn!n_1!n_2!dots n_r!$ regardless whether you care about the order of the groups. It cannot be the case that you divide by $r!$, since for example when $n=3,n_1=1,n_2=2$, then $frac3!1!2!=3$ is not divisible by $2!$.
If all groups are the same size, the you do need to add $r!$ to the denominator to discount for order, so the formula is $fracn!r!(n/r)!^r$.
In general, suppose the numbers $n_i$ are partitioned into several groups, where the $k^th$ group has size $lambda_k$, and the $n_i=n_j$ if and only if $n_i$ and $n_j$ are in the same group. Then the number of ways to partition the objects into groups of size $n_1,dots,n_r$ without respect to order is
$$
fracn!left(prod_i=1^r n_i!right)left(prod_k=1^m lambda_j!right).
$$
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
edited Apr 2 at 14:17
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
answered Apr 2 at 4:02
Mike EarnestMike Earnest
28.2k22152
28.2k22152
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
add a comment |
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
$begingroup$
Of course, you are right.
$endgroup$
– Jair Taylor
Apr 2 at 5:16
add a comment |
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$begingroup$
That's correct.
$endgroup$
– Jair Taylor
Apr 2 at 1:40
$begingroup$
Does the denominator need to add up to n in the first case?
$endgroup$
– Sebastian
Apr 2 at 2:13
$begingroup$
Yes, you need $sum_i=1^r n_i = n$ since the sizes of the blocks must add up to the size of the set
$endgroup$
– Jair Taylor
Apr 2 at 2:41