How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be a set?

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

2xy/9 is the answer to this problem