Question

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

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

Answer 1

Answer: (B)

$\frac{n(n-4)(n-5)}{3!}$

Choose one object first. Then think of the remaining $n-3$ objects in a line. There are then $n-4$ paces and choose the object to the left of the left space and right of the right space.

This last step is in 1-1 correspondence with the ways to have the final $2$ objects; therefore we divide by $3$ (corresponding to the initial first choice, which could have been any of the $3$ objects to start) giving

$\frac{n(n-4)(n-5)}{6}$