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)

Given, number of persons $=7$
Now, treating the two persons as one, we have 6 persons.
Total number of sitting arrangements $=7$ !
Favourable number of arrangements $=6$ !
2 persons can be arranged in two ways.
Total number of favourable arrangments $=2(6!)$
$\therefore$ Required probability $=\frac{2(6!)}{7!}=\frac{2}{7}$