How many $3$ integer subsets have no consecutive integers, where integers are less than $20$? 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.
 A: This is a simple way
Consider 20 unlabelled counters (C). Take away 3, now 17 are left, and there are 18 gaps (including ends) where we can insert back the 3 to comply with non-consecutive stipulation. Now serially number the counters.
_C_C_C_C_C_C_C_C_C_C_C_C_C_C_C_C_C_
$$\text{thus there can be}\;\; {18\choose3} = 816\;\; \text{such subsets}$$ 
A: We can represent the 3 integers by 3 dots, 
and then let $x_i$ for $1\le i\le 4$ be the number of integers in the 4 gaps created by the dots.
Then $x_1+x_2+x_3+x_4=17$ where $x_1, x_4\ge0$ and $x_2, x_3\ge1$;
so if we let $y_i=x_i$ for $i=1,4$ and $y_i=x_i-1$ for $i=2,3$, 
we get $y_1+y_2+y_3+y_4=15$ where $y_i\ge0$ for each $i$, 
and this equation has $\dbinom{18}{3}=816$ nonnegative integer solutions.

$\textbf{Alternate solution}$:
Line up 3 red dots (representing the integers to be chosen), and put aside
2 dots (which will be blockers);
this leaves 15 dots and 3 red dots to be arranged in a row, and this can be done in 
$\dbinom{18}{3}=816$ ways.
(Now insert the blockers between the red dots, so that they are not consecutive.)
A: 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:
$$\binom{18}3$$
A: It's probably easier to count how many actually have two consecutive integers and subtract that from the total number of three element subsets.
First we note that there are  $18$ subsets with all elements consecutive.
Now we wish count the number of three element subsets such that exactly two elements are consecutive. First, let's count the number of these subsets which additionally have the non-consecutive element as the largest. If the consecutive elements are $n-1$ and $n$, then the remaining element can be any of the numbers $n+2,n+3,\dots,20$. Since the lowest $n$ can be is $2$, we get
$$17+16+\dots+1=\frac{17\times18}{2}=153$$
subsets of this form. By symmetry we also have $153$ subsets which have the non-consecutive element as the smallest. 
This gives us a total of $18+153+153=324$ subsets with two consecutive elements.
The total number of subsets is $\binom{20}{3}=\frac{20\times19\times18}{6}=1140$, so by subtraction we find the total number with no consecutive elements is $1140-324=816$.
A: We can apply the Inclusion-Exclusion Principle to determine the number of three-element subsets of the set $\{1, 2, 3, \ldots, 20\}$ that do not contain consecutive integers.  
There are $\binom{20}{3}$ subsets with three elements.  From these, we must subtract those subsets that contain at least two consecutive numbers.  There are $19$ pairs of consecutive numbers since the smaller number in the pair can be any other number other than $20$ and $18$ ways of selecting the third element of the set, so there are $19 \cdot 18$ three element subsets that contain two consecutive numbers. However, we have counted those subsets that contain three consecutive numbers twice, once when we selected the two smallest numbers in the subset and once when we selected the two largest numbers in the subset.  There are $18$ sets of three consecutive numbers in the set since the smallest number in the set cannot be $19$ or $20$.  Hence, by the Inclusion-Exclusion Principle, the number of three-element subsets of the set $\{1, 2, 3, \ldots, 20\}$ that do not contain consecutive numbers is 
$$\binom{20}{3} - 19 \cdot 18 + 18 = 1140 - 342 + 18 = 816$$ 
