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Mismatched trade confirmations

CombinatoricsMedium

$5$ trade confirmations belong to $5$ different counterparties, and each is placed into one of $5$ folders, one per folder, all $120$ arrangements equally likely. Folder $k$ is the correct one for counterparty $k$ . What is the probability that at least $1$ confirmation lands in its correct folder?

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Hint

Count the complementary event, where no confirmation is in its correct folder.

Solution

The number of arrangements with no match is $5! (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}) = 44$ .
The probability of no match is $\frac{44}{120} = \frac{11}{30}$ .
The probability of at least one match is $1 - \frac{11}{30}$ .
The answer is $\frac{19}{30}$ .

Answer: $\frac{19}{30}$

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