[Math] $20$ people sit at a round table, how many ways can we choose $3$ with no $2$ being neighbors

combinationscombinatoricsdiscrete mathematics

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?

Best Answer

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 $= \frac{20}{17}\times\binom{17}{3} = 800$