$20$ 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.

  • $\begingroup$ 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. $\endgroup$
    – Lourrran
    Commented Feb 15, 2023 at 12:40

1 Answer 1


The logic for the second formula is:

  • 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


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