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Dear Marina,

I have been asked this question before via pm. See below my answer.

It seems like the factor in front of (k 3) should be 3 and not 2, as there are 3 possible arrangements as you correctly pointed out. As you also remarked, a constant factor does not really matter though...

Perhaps somehow one can show that when the condition holds for 2 out of the 3 arrangements of a triplet (i,j,k), the third condition holds automatically. If that would be the case, one could indeed write the “2”.
I gave it a quick try, but I could not confirm this conjecture, so it is at the very least not trivial (or maybe simply wrong, and it’s just a typo in the sample solution).


Best,

Thomas