# Finding logic for a different form of an answer.

$$20$$ persons are sitting on a clean round table. Find the number of ways of selecting $$4$$ persons. Such that no two of them are consecutive.

My approach

I selected one out of $$20$$, and the remaining $$3$$ should be selected out of $$17$$ persons (removing the selected one and his adjacents). With at least gap of one person between them. So this is now equivalent to selecting $$3$$ person in a linear arrangement not adjacent to each other from $$17$$ persons for which the ways is $$^{15}\text{C}_3$$.

Multiplying both selection gives, $$^{20}\text{C}_1\cdot^{15}\text{C}_3$$

But this has repetition four times for each of the first selected person.

So, total cases is $$\frac{^{20}\text{C}_1\cdot^{15}\text{C}_3}{4}$$

But the text book in which the question was gives answer as $$\frac{^{20}\text{C}_1\cdot^{15}\text{C}_3}{4}$$ and $$^{17}\text{C}_4- ^{15}\text{C}_2$$

And has a note that both answers have separate logic. Obviously these are inter convertible and has same value, but I am unable to find a different logic for this one.

• We select 1 person, example person at seat n°10. So n°9 and n°11 are forbidden for the rest of the process. If person n°2 is seat n°12, impact for 3rd person is not the same as if personn°2is at seat n°13. If the 2 first persons selected are n°10 and n°12, only n°9, n°11 and n°13 are forbidden for te following steps. n°11 is duplicated. Another way to see this is to select 10 persons, with same constraint. With your method, you find $0$ possibility ; in reality, there are $2$ possibilities. Commented Feb 15, 2023 at 12:40

• Firstly seat in a straight line. Imagine $$16$$ empty chairs laid out in a line. There will be $$17$$ interstices (including two at the ends) so $$4$$ non-adjacent chairs can be inserted into the line in $$\binom{17}4$$ ways
• But this includes an arrangement in which two of the four chairs are at the extremities (so would become adjacent if rolled into a circle). So now do a similar straight line exercise with $$2$$ chairs to be inserted in $$16$$ empty chairs using only the $$15$$ interior interstices in $$\binom{15}2$$ ways, and subtract