$20$ people sit at a round table, how many ways can we choose $3$ with no $2$ being neighbors? My thought process to this problem was as follows:
$1st$ move you have $20$ choices, when you pick you eliminate $3$ people, the first person and their two neighbors. The $2nd$ move you have $20-3$ people to choose from so $17$. And so on.
Does: $20 \cdot 17 \cdot 14$ make sense as an answer?
 A: Another way: 
Looking clockwise, attach a "general" person to each of $3$ "special" persons $\fbox{SG}$
There are now $3$ boxes + $14$ individuals $= 17$ entities
Place the boxes in $\binom{17}{3}$ ways,
but since you are allowing each entity only $17$ starting points instead of $20$,
multiply by $\frac{20}{17}$ to get ans $= \frac{20}{17}\times\binom{17}{3} = 800$ 
A: There are $20$ ways to choose $3$ people sitting in three consecutive seats. There are $20\cdot16$ ways to choose $3$ people where $2$ are sitting together with the third sitting apart from them. Thus we have $17\cdot20=340$ choices that do not qualify. Excluding these from the total, the number of ways to choose $3$ with no $2$ being neighbors is:
$${20\choose3} - 340 = 800$$
A: Call the oldest person at the table $1$, the person to his left $2$, and so forth until you get around to $1$'s immediate right, which is $20$.  Now order each qualifying triple $A,B,C$ such that $1 \leq A < B < C \leq 20$ to count each one exactly once.
Starting with $A=1$, we can have $(1,3,5), (1,3,6), ..., (1,3,19), (1,4,6), ..., (1, 17, 19),$ which by my counting gives $T_{15} = 120$ ($T_n$ is the triangular number).
The case $A=2$ is the same because we can now go up to $20$.
The case $A=3$ gives $T_{14} = 105$ choices.  The case $A=4$ gives $T_{13} = 91$ choices.  Continuing the pattern to the end, $A=16$ gives $T_1 = 1$ choice.
Adding all of these up we find the answer is
$$S = T_{15} + \sum_{n=1}^{15} T_n = 800.$$
