Question

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{2}{7}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Total seven persons are seated in a row in ${}^7{P}_1$ ways. Two particular persons are seated next to each other, i.e., we consider two persons as a single unit.
$\therefore$ These two persons and rest five persons are seated in a row by $6!$ ways but these two persons can be arranged in $2!$ ways.
$\therefore$ Required probability $=\frac{2!\times 6!}{{}^7{P}_1}$
$=\frac{2!\times 6!}{7!}$
$=\frac{2}{7}$