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Counting triplets where each elements are from different groups

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

0

1

$begingroup$

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

share|cite|improve this question

asked 22 hours ago

someone123123someone123123

455414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, there can be N>3 groups
  $endgroup$
  – someone123123
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  14 hours ago
  

  
  
  
  

add a comment | 

0

1

$begingroup$

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

share|cite|improve this question

asked 22 hours ago

someone123123someone123123

455414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, there can be N>3 groups
  $endgroup$
  – someone123123
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  14 hours ago
  

  
  
  
  

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

share|cite|improve this question

asked 22 hours ago

someone123123someone123123

455414

$endgroup$

share|cite|improve this question

asked 22 hours ago

someone123123someone123123

455414

share|cite|improve this question

asked 22 hours ago

someone123123someone123123

455414

asked 22 hours ago

someone123123someone123123

455414

asked 22 hours ago

someone123123someone123123

455414