Question

If $n$ objects are arranged in a row, the number of ways of selecting three of these objects so that no two adjacent objects are selected is

• A. ${}^{n-2}{C}_3$
• B. ${}^{n-2}{C}_{n-5}$
• C. ${}^{n-3}{C}_2$
• D. ${}^{n-3}{C}_{n-5}$

Answer 1

Answer: (A), (B)

The correct options are
A ${}^{n-2}{C}_3$
B ${}^{n-2}{C}_{n-5}$

A and B

Let A B and C be the 3 objects selected and $x_1,x_2,x_3,x_4$ be the number of objects on either sides of the selected objects as shown below.

$x_{1}A{x}_{2}B{x}_{3}C{x}_4$

we have

$x_1{+x}_2+x_3+x_4=n-3$ where $x_1,x_4\ge 0,x_2,x_3\ge 1$

So required number of ways

= coefficient of $x^{n-3}$ in ${(1+x+x^2+x^3+....)}^2{(x+x^2+x^3+....)}^2$

= coefficient of $x^{n-5}$ in ${(1+x+x^2+x^3+....)}^4$

= coefficient of $x^{n-5}$ in ${(1-x)}^{-4}$ =${{}^{n-5+4-1}C}_{n-5}={{}^{n-2}C}_{n-5}={{}^{n-2}C}_3$