Counting triplets where each elements are from different groups Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Recursive and closed form solution for choosing $n$ pairs/triplets.. of $kn$ elements.Counting triplets with red edges in each pairSelect elements from $N$ setsTwo different results for splitting sample points into groups of a certain size?Counting distinct positive valued k-tuples that sum to n where each entry can be no greater than some value.Arranging m distinct groups of k elements each?How many ways can numbers be split into different groupsForming triplets from pairsCounting different way of arranging letters from two groups, without repetitionsThe number of sequences of length $10$ that can be formed with $5$ unique digits containing two of each where no two adjacent elements are alike
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Counting triplets where each elements are from different groups
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Recursive and closed form solution for choosing $n$ pairs/triplets.. of $kn$ elements.Counting triplets with red edges in each pairSelect elements from $N$ setsTwo different results for splitting sample points into groups of a certain size?Counting distinct positive valued k-tuples that sum to n where each entry can be no greater than some value.Arranging m distinct groups of k elements each?How many ways can numbers be split into different groupsForming triplets from pairsCounting different way of arranging letters from two groups, without repetitionsThe number of sequences of length $10$ that can be formed with $5$ unique digits containing two of each where no two adjacent elements are alike
$begingroup$
Let's say we have given groups with more elements each, is there easy way to count number of ways to choose three elements from different groups.
For example if our groups are $1, 2 text and 3, 4$. There are two ways to form the triplets $1, 2, 3, 1, 2, 4$.
I tried fixing all three groups but that works in $O(N^3)$, where $N$ is the number of groups. Is it possible to make it in $O(N)$?
combinatorics
asked Mar 27 at 7:04
someone123123someone123123
464415
$endgroup$
add a comment |
$begingroup$
Let's say we have given groups with more elements each, is there easy way to count number of ways to choose three elements from different groups.
For example if our groups are $1, 2 text and 3, 4$. There are two ways to form the triplets $1, 2, 3, 1, 2, 4$.
I tried fixing all three groups but that works in $O(N^3)$, where $N$ is the number of groups. Is it possible to make it in $O(N)$?
combinatorics
asked Mar 27 at 7:04
someone123123someone123123
464415
$endgroup$
$begingroup$
Do you mean there can be $N>3$ groups? (Because if there are exactly $N=3$ groups the answer is trivial. But then I'm surprised your example has $N=3$.)
$endgroup$
– antkam
Mar 27 at 14:50
$begingroup$
Yes, there can be N>3 groups
$endgroup$
– someone123123
Mar 27 at 14:55
$begingroup$
Can I assume all the groups are disjoint, i.e. no element appears in two or more groups? If so, I have an $O(N)$ solution... Is this homework? If so I will give you a hint, otherwise I will post the solution.
$endgroup$
– antkam
Mar 27 at 15:33
add a comment |
$begingroup$
Let's say we have given groups with more elements each, is there easy way to count number of ways to choose three elements from different groups.
For example if our groups are $1, 2 text and 3, 4$. There are two ways to form the triplets $1, 2, 3, 1, 2, 4$.
I tried fixing all three groups but that works in $O(N^3)$, where $N$ is the number of groups. Is it possible to make it in $O(N)$?
combinatorics
asked Mar 27 at 7:04
someone123123someone123123
464415
$endgroup$
Let's say we have given groups with more elements each, is there easy way to count number of ways to choose three elements from different groups.
For example if our groups are $1, 2 text and 3, 4$. There are two ways to form the triplets $1, 2, 3, 1, 2, 4$.
I tried fixing all three groups but that works in $O(N^3)$, where $N$ is the number of groups. Is it possible to make it in $O(N)$?
combinatorics
combinatorics
asked Mar 27 at 7:04
someone123123someone123123
464415
asked Mar 27 at 7:04
someone123123someone123123
464415
asked Mar 27 at 7:04
someone123123someone123123
464415
asked Mar 27 at 7:04
someone123123someone123123
464415
asked Mar 27 at 7:04
someone123123someone123123
464415
464415
$begingroup$
Do you mean there can be $N>3$ groups? (Because if there are exactly $N=3$ groups the answer is trivial. But then I'm surprised your example has $N=3$.)
$endgroup$
– antkam
Mar 27 at 14:50
$begingroup$
Yes, there can be N>3 groups
$endgroup$
– someone123123
Mar 27 at 14:55
$begingroup$
Can I assume all the groups are disjoint, i.e. no element appears in two or more groups? If so, I have an $O(N)$ solution... Is this homework? If so I will give you a hint, otherwise I will post the solution.
$endgroup$
– antkam
Mar 27 at 15:33
add a comment |
$begingroup$
Do you mean there can be $N>3$ groups? (Because if there are exactly $N=3$ groups the answer is trivial. But then I'm surprised your example has $N=3$.)
$endgroup$
– antkam
Mar 27 at 14:50
$begingroup$
Yes, there can be N>3 groups
$endgroup$
– someone123123
Mar 27 at 14:55
$begingroup$
Can I assume all the groups are disjoint, i.e. no element appears in two or more groups? If so, I have an $O(N)$ solution... Is this homework? If so I will give you a hint, otherwise I will post the solution.
$endgroup$
– antkam
Mar 27 at 15:33
$begingroup$
Do you mean there can be $N>3$ groups? (Because if there are exactly $N=3$ groups the answer is trivial. But then I'm surprised your example has $N=3$.)
$endgroup$
– antkam
Mar 27 at 14:50
$begingroup$
Do you mean there can be $N>3$ groups? (Because if there are exactly $N=3$ groups the answer is trivial. But then I'm surprised your example has $N=3$.)
$endgroup$
– antkam
Mar 27 at 14:50
$begingroup$
Yes, there can be N>3 groups
$endgroup$
– someone123123
Mar 27 at 14:55
$begingroup$
Yes, there can be N>3 groups
$endgroup$
– someone123123
Mar 27 at 14:55
$begingroup$
Can I assume all the groups are disjoint, i.e. no element appears in two or more groups? If so, I have an $O(N)$ solution... Is this homework? If so I will give you a hint, otherwise I will post the solution.
$endgroup$
– antkam
Mar 27 at 15:33
$begingroup$
Can I assume all the groups are disjoint, i.e. no element appears in two or more groups? If so, I have an $O(N)$ solution... Is this homework? If so I will give you a hint, otherwise I will post the solution.
$endgroup$
– antkam
Mar 27 at 15:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT.
This assumes the groups are disjoint, i.e. no element appears in multiple groups.
It turns out to be easier to count the complement. You pick $3$ elements from the entire universe.
How many $3$-subsets have all elements from the same group? This can be counted easily in $O(N)$.
How many $3$-subsets have all elements from only two different groups? This can be counted easily in $O(N^2)$ and with a small trick can also be counted in $O(N)$.
answered Mar 28 at 12:45
antkamantkam
3,214412
$endgroup$
add a comment |
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$begingroup$
HINT.
This assumes the groups are disjoint, i.e. no element appears in multiple groups.
It turns out to be easier to count the complement. You pick $3$ elements from the entire universe.
How many $3$-subsets have all elements from the same group? This can be counted easily in $O(N)$.
How many $3$-subsets have all elements from only two different groups? This can be counted easily in $O(N^2)$ and with a small trick can also be counted in $O(N)$.
answered Mar 28 at 12:45
antkamantkam
3,214412
$endgroup$
add a comment |
$begingroup$
HINT.
This assumes the groups are disjoint, i.e. no element appears in multiple groups.
It turns out to be easier to count the complement. You pick $3$ elements from the entire universe.
How many $3$-subsets have all elements from the same group? This can be counted easily in $O(N)$.
How many $3$-subsets have all elements from only two different groups? This can be counted easily in $O(N^2)$ and with a small trick can also be counted in $O(N)$.
answered Mar 28 at 12:45
antkamantkam
3,214412
$endgroup$
add a comment |
$begingroup$
HINT.
This assumes the groups are disjoint, i.e. no element appears in multiple groups.
It turns out to be easier to count the complement. You pick $3$ elements from the entire universe.
How many $3$-subsets have all elements from the same group? This can be counted easily in $O(N)$.
How many $3$-subsets have all elements from only two different groups? This can be counted easily in $O(N^2)$ and with a small trick can also be counted in $O(N)$.
answered Mar 28 at 12:45
antkamantkam
3,214412
$endgroup$
HINT.
This assumes the groups are disjoint, i.e. no element appears in multiple groups.
It turns out to be easier to count the complement. You pick $3$ elements from the entire universe.
How many $3$-subsets have all elements from the same group? This can be counted easily in $O(N)$.
How many $3$-subsets have all elements from only two different groups? This can be counted easily in $O(N^2)$ and with a small trick can also be counted in $O(N)$.
answered Mar 28 at 12:45
antkamantkam
3,214412
answered Mar 28 at 12:45
antkamantkam
3,214412
answered Mar 28 at 12:45
antkamantkam
3,214412
answered Mar 28 at 12:45
antkamantkam
3,214412
3,214412
add a comment |
add a comment |
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$begingroup$
Do you mean there can be $N>3$ groups? (Because if there are exactly $N=3$ groups the answer is trivial. But then I'm surprised your example has $N=3$.)
$endgroup$
– antkam
Mar 27 at 14:50
$begingroup$
Yes, there can be N>3 groups
$endgroup$
– someone123123
Mar 27 at 14:55
$begingroup$
Can I assume all the groups are disjoint, i.e. no element appears in two or more groups? If so, I have an $O(N)$ solution... Is this homework? If so I will give you a hint, otherwise I will post the solution.
$endgroup$
– antkam
Mar 27 at 15:33