How many ways to select $4$ people at a round table such that no two of the selected people are adjacent? 
There are $100$ people sited in a circular table with $100$ chairs. How
many ways to select $4$ such that each chosen person is not adjacent to
any other chosen person.


The problem is equivalent to the following

Let there be $100$ empty chairs in a circular table. How many ways to choose $4$ chairs such that each chosen chair is not adjacent to any other chair

When it's a straight table, we can remove a chair between each chosen one yielding $\binom{97}{4}$ by bijection argument. It can maybe be applied when the table is circular.

I found a solution using PIE
$$\binom{100}{4}-\frac{100\binom{98}{2} \cdot 4-100 \cdot 9 7 \cdot 6+4 \cdot 100}{4}= 3460375$$
 A: From the straight table case, subtract the case when both #1 and #100 are selected. When the ends of row are joined to form a circle, we get our desired arrangement.
By the bijection argument, it's $$\binom{97}{4} - \binom{95}{2}$$

 Since it remains to select two non-adjacent chairs from #3 to #98 inclusive ie, 96 chairs.

A: I will answer the part which you say is "equivalent to" the question you want to ask because the original question says $100$ people seated in $100$ chairs. Also when "people" implies distinct entities whereas in the "equivalent" question they are faceless.
If we cut the circle to straighten it out $X ..................Y$, two cases arise: (i) we use neither $X$ nor $Y\;$ (ii) we use one of $X$ or $Y$
in either case, we have $95$ slots available,
thus the number of ways $= \dbinom{95}4 + 2\dbinom{95}3 = 3460375 $

Extra Answer
You may find this answer even simpler.
$1.$ Remove $4$ chairs and place them in the $96$ interstices.
$2.$ The chairs have only $96$ "starting" places instead of the actual $100$, so multiply by $\frac{100}{96}$
Thus $\dbinom{96}4\dfrac{100}{96} = 3460375$
