There are
total possible ways to pick any three integers from the set.
Of the total, there are 18 consisting of consecutive triplets
.
Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs
, and for each pair 18 possible choices for the third integer (for instance,
can be taken with 3, 4, ..., 20), to a total of
. To avoid double-counting (e.g.
can't go with 3;
can't go with 1 or 4), we subtract 1 from the extreme pairs
and
(twice), and 2 from the rest (17 times).
So, the number of triplets that don't consist of pairwise consecutive integers is
I don't know how useful this would be to you, but I've verified the count in Mathematica:
In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816
