Seven persons are to be seated in a row.

Solution:

C. 2/7

Explanation:

Given that 7 persons are to be seated in a row.

 

If two persons sit next to each other, then consider these two persons as 1 group.

Now we have to arrange 6 persons.

$\therefore$ Number of arrangement $=2 ! \times 6 !$

Total number of arrangement of 7 persons $=7 !$

Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$

$\therefore$ Required probability $=\frac{2 ! \times 6 !}{7 !}$

$=\frac{2 \times 1 \times 6 !}{7 \times 6 !}$

$=\frac{2}{7}$

Hence, the correct option is (C).

= 2/7

Hence, the correct option is (C)