Question

$n$ different things are arranged around a circle, number of ways of selecting $3$ objects when no two selected objects put together, is

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

Answer 1

Answer: (B)

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