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I have 2 lists of numbers A = Range[16, 65] and B = Range[10, 16]. I want to form all possible combinations $(a,b)$ of the elements in those lists such that $a - b > 5$. I can get what I want with:

Tuples[{A, B}] /. {a_, b_} /; a - b <= 5 -> Nothing

How can I get the same list using Table?

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    $\begingroup$ Table[] with If[] and Nothing ought to do it. $\endgroup$ – J. M. is in limbo♦ Jul 25 '16 at 1:42
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Table[{a, b}, {a, 16, 65}, {b, 10, Min[16, a - 5 - 1]}] // Flatten[#, 1] &
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  • $\begingroup$ rather Table[{a, b}, {a, 16, 65}, {b, 10, Min[16, a - 5]}] // Flatten[#, 1]&... $\endgroup$ – amrods Jul 25 '16 at 1:56
  • $\begingroup$ @amrods Thanks! But I think it should be Min[16, a - 5 - 1] because a - b > 5 means b <= a - 6 $\endgroup$ – wuyingddg Jul 25 '16 at 2:04
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To do it with Table modify your Tuples example:

Flatten[Table[If[a - b > 5, {a, b}, Nothing],
   {a, Range[16, 65]}, {b, Range[10, 16]}], 1];

If you are not committed to Table another idea is to use Outer

Flatten[Outer[If[#1 - #2 > 5, List[#1, #2], Nothing] &, 
   Range[16, 65], Range[10, 16]], 1];
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Use Pick

Block[{t=Tuples[{A,B}]},Pick[t,-UnitStep[Apply[Subtract,t,{2}]+5],1]]
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    $\begingroup$ This does not appear to use Table, as requested in the question. $\endgroup$ – bbgodfrey Jul 25 '16 at 4:01
  • $\begingroup$ @bbgodfrey oh, yes...... $\endgroup$ – Wjx Jul 25 '16 at 4:05

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