Question

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

Answer 1

There are $\dbinom{20}3=1140$ total possible ways to pick any three integers from the set.

Of the total, there are 18 consisting of consecutive triplets $(\{1,2,3\},\{2,3,4\},\ldots,\{18,19,20\})$.

Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs $(\{1,2\},\{2,3\},\ldots,\{19,20\})$, and for each pair 18 possible choices for the third integer (for instance, $\{1,2\}$ can be taken with 3, 4, ..., 20), to a total of $19\times18=342$. To avoid double-counting (e.g. $\{1,2\}$ can't go with 3; $\{2,3\}$ can't go with 1 or 4), we subtract 1 from the extreme pairs $\{1,2\}$ and $\{19,20\}$ (twice), and 2 from the rest (17 times).

So, the number of triplets that don't consist of pairwise consecutive integers is

$1140-(18+342-(2\times1+17\times2))=816$

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