If '0-7' has less number of picks compared to '1-6', then we will have no solution at all. But, it is possible that ('0-7','1-6') has (3,2) picks or even (2,2) picks respectively. In the first case there will be exactly 1 selection from 0 and 7 in any solution. In the 2nd case, 0 and 7 will not be selected in any solution.
Re^2: Multiple Combinatorics
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Care to clarify that? Maybe add some examples? Cos like JavaFan, I don't get it.
As soon as you pick any two values from group 2 (1..6), you've also already picked two values from group 1 (0..7), so there is no way to comply with your "every resulting row should satisfy the initial condition" requirement.
You would either a) a pick two from both groups and therefore have 4 from the larger group; or b) pick nothing from the larger group, in which case you have nothing from the larger group.
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if Group B (1..6) is subset of Group A(0..7) and Pick(B)=2 is equal to Pick(A)=2 then we apply 'b): pick nothing from larger group';
In this case, we still have picked 2 values from larger group automatically, which holds the initial condition. Only thing that '0' and '7' won't be picked at all in any resultset. ( from the given data).