Question

$n(\ge 5)$ persons are sitting in a row. Three of these are selected at random, the probability that no two of the selected persons are together is

• A. $\frac{{}^{n-3}{P}_2}{{}^n{P}_2}$
• B. $\frac{{}^{n-3}{C}_2}{{}^n{C}_2}$
• C. $\frac{(n-3)(n-4)}{n(n-1)}$
• D. $\frac{{}^{n-3}{C}_2}{{}^n{P}_2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{{}^{n-3}{C}_2}{{}^n{C}_2}$
B $\frac{(n-3)(n-4)}{n(n-1)}$
C $\frac{{}^{n-3}{P}_2}{{}^n{P}_2}$

Select $3$ persons (resulting in $2$ gaps in between them and $2$ in the extremes i.e. $4$ gaps in all). Place $1$ person each in these gaps so that the $3$ persons are not consecutive. The remaining $(n-5)$ persons can be distributed in the $4$ gaps in $C_{4-1}^{n-5+4-1}=C_3^{n-2}=\frac{(n-2)(n-3)(n-4)}{6}$ ways.

Hence, probability = $\frac{C_3^{n-2}}{C_3^n}=\frac{\frac{(n-2)(n-3)(n-4)}{6}}{\frac{n(n-1)(n-2)}{6}}=\frac{(n-3)(n-4)}{n(n-1)}$