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

The correct option is D $\frac{2}{7}$
Total number of persons $=7$

Number of seats $=7$

We know $‘n^{\prime }$ persons can be arranged in

$\text{'}r\text{'}$ seats in ${}^n{P}_r$ ways where ${}^n{P}_r=\frac{n!}{(n-r)!}(n\ge r)$

$\therefore 7$ persons can be arranged in $7$ seats in

${}^7{P}_{7}ways=\frac{7!}{(7-7)!}$

$=\frac{7!}{0!}$

$=7!$ ways

If these two persons are considered as a single group, then number of units $=6$

$6$ persons can be arranged in $6$ seats in $6!$ ways.

Those $2$ persons can be arranged among themselves in $2!$ ways.

$\therefore$ No. of favorable cases $=2!\times 6!=2\times 6!$

Probability$=\frac{Numberoffavorablecases}{totalnumberofcases}$

$=\frac{2\times 6!}{7!}=\frac{2\times 6!}{7\times 6!}=\frac{2}{7}$

Final Answer: C