Question

$n$ different things are arranged around a circle. Find the number of ways of selecting $3$ objects such that no two selected objects are together.

• A. $\frac{(n-3)(n-4)}{3!}$
• B. $\frac{n(n-4)(n-5)}{3!}$
• C. $\frac{n(n-5)}{6}$
• D. $\frac{n(n-4)}{3}$

Answer 1

The correct option is B $\frac{n(n-4)(n-5)}{3!}$

Let the objects are denoted by $A_1,A_2,A_3......A_n$arranged in a circle, we have to select 3 object so that no two of them are together. To do so, we find the ways in which two or three object together, now the ways in which two or three are together is obtained in the following manner with $A_1$. The number such triplets is $A_1,A_2,A_3;A_1,A_2{A}_4....A_1,A_2,A_{n-1}$
so the number of same kind of triplets when we first take $A_1$ is $(n-3)$ and similarly the others
But total number of triplets are ${}^n{C}_3$
$\therefore$ Required such ways ${=}^n{C}_3-n(n-3)$
where $(n-3)$ are total number of triplets
$=\frac{n(n-1)(n-2)}{6}-n(n-3)$
$=\frac{n(n-4)(n-5)}{6}$
$=\frac{n(n-4)(n-5)}{3!}$