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Theoretical side

Simple example: If I have two sets $A_1=\{1,3\} ,A_2=\{2,3\}$

and Permutations[{1, 2, 3}]= $\left( \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 1 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \\ 3 & 2 & 1 \\ \end{array} \right)$

Solution steps

1- Choose the rows whose sets $A_1=\{1,3\} ,A_2=\{2,3\}$ are partial and then count the number of rows $D_2=\{1,3,2\},\{3,1,2\},\{2,3,1\},\{3,2,1\}$

$S_2=Count[D_2]/Count[Permutations[\{1, 2, 3\}]]=4/6$

2- Delete $D_2$ from $Permutations[\{1, 2, 3\}]$ and choose from the triple order whose sets $A_1=\{1,3\} ,A_2=\{2,3\}$ are partial and then count the number of rows

$D_3=\{1,2,3\},\{2,1,3\}$

$S_3=Count[D_3]/Count[Permutations[\{1, 2, 3\}]]=2/6$

How can you write code that achieves this algorithm and maintains globality?

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#Edit

sets are partial:The group is contained within the second order

$D_2$:Such that

$A_1=\{1,3\}\subset\{1,3,-\}$ take $\{1,3,-\}$ $A_1=\{1,3\}\subset\{3,1,-\}$ take $\{3,1,-\}$ $A_2=\{2,3\}\subset\{2,3,-\}$ take $\{1,3,-\}$ $A_2=\{2,3\}\subset\{3,2,-\}$ take $\{3,1,-\}$

Then; We obtain $$D_2=\{1,3,2\},\{3,1,2\},\{2,3,1\},\{3,2,1\}$$

$D_3$:Such that

$A_1=\{1,3\}\subset\{1,2,3\}$ take $\{1,2,3\}$ $A_1=\{2,3\}\subset\{2,1,3\}$ take $\{2,1,3\}$

Then; We obtain $$D_3=\{1,2,3\},\{2,1,3\}$$

Improve this question
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  • $\begingroup$ Please clarify what you mean by "sets are partial". $\endgroup$
    – Carl Woll
    May 10, 2022 at 0:34
  • $\begingroup$ @CarlWoll sets are partial: The set is contained within the second order $D_2$; $A_1=\{1,3\}\subset\{1,3,-\}$ or $A_1=\{1,3\}\subset\{3,1,-\}$ And, The set is contained within the thirds order $D_3$; $A_1=\{1,3\}\subset\{1,2,3\}$ $\endgroup$
    – Emad kareem
    May 10, 2022 at 0:46

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