The captains of five cricket teams, including India and Australia, are lined up randomly next to one other for a group photo. What is the probability that the captains of India and Australia will stand next to each other?
Step-by-step Solution:
Five captains are arranged randomly in a line. The total number of possible orderings is \[ 5! = 120. \] Treat the India and Australia captains as a single block to ensure they stand next to each other. Then we have this block plus the other three captains, i.e. \(4\) items, which can be arranged in \(4!\) ways. Inside the block the two captains can be ordered in \(2!\) ways. Hence the number of favourable arrangements is \[ 2!\cdot 4! = 2\cdot 24 = 48. \] Therefore the required probability is \[ \frac{48}{120}=\frac{2}{5}. \] \[ \boxed{\dfrac{2}{5}} \]