TITLE: How many $3$ integer subsets have no consecutive integers, where integers are less than $20$?
QUESTION [5 upvotes]: I have to determine how many integers between $1$ and $20$ are possible if no two consecutive integers are in a set. I've thought it has something to do with a combination of an element $(a,a+2,a+4)$ and all of these possibilities but I know I'm missing something. I don't know how I would determine all of the possibilities, especially say $a=19$, would $a+2$ loop back around to $1$? I may be completely off in this approach though, so any help would be great.

REPLY [2 votes]: You want three numbers from $1$ to $20$, such that no two are consecutive. Call them $a$, $b$, and $c$; without loss of generality, $a<b<c$.
Note that, if $a<b$, and they're not consecutive, we can write $a+1<b$. Using this observation, we see you want the number of ways of selecting $a$, $b$, and $c$ from among the integers, such that:
$$(1\le a)\&(a+1<b)\&(b+1<c)\&(c\le 20)$$
In other words, you want the number of ways of selecting three integers, $a$, $b$, and $c$, such that:
$$1\le a<b-1<c-2\le 18$$
This means that, the question is equivalent of asking how many ways we can choose $3$ integers ($a$, $b-1$, and $c-2$) between $1$ and $18$. In other words, we want to know in how many ways we can choose three objects out of eighteen objects. Thus, the answer is:
$$\binom{18}3$$