"be consistent"  
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For a hand with a 5 to avoid scoring 2 points, it must have no pair, and no subset that sums to 15. Since that means we can only have one of each of the pairs (1, 9), (2, 8), (3, 7), (4, 6) that sum to 10, the hands to consider consist of the 16 ways we can choose one from each of those pairs. If we have 9 we cannot have 6, so we must have 4, so we cannot have 2, so we must have 8, so we cannot have 7, so we must have 3. But 3+4+8 = 15, so we cannot have 9. So we must have 1; if we have 8 we cannot have 7, or 6, so we must have 3 and 4. So we have 3+4+8 = 15 again, which means we cannot have 8. So we must have 1 and 2, which means we cannot have 7, so we must have 3. But 1+2+3+5+4 = 15, and 1+3+5+6 = 15. So there is no hand including a five which doesn't include at least 2 points. Hugo In reply to Re: Generating all 5card hands
by hv

