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Q.
Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is
Probability
Solution:
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 $= \frac{2!\times6!}{^{7}P_{1}}$
$ = \frac{2!\times6!}{7!} $
$= \frac{2}{7}$