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Q: A large number of drunk guests arrive at a hotel where they have booked specific rooms. A careless receptionists hands over keys at random. What is the probability that at least one guest ends up in a room she booked?

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A: A useful concept to understand is that of Derangement. A derangement is the number of ways a set can be permuted such that none of the elements are in their respective positions. For example if \(\{a,b,c\}\) is a set, then the derangements of the set are

\(\{b,c,a\}\)\(\{c,a,b\}\)The above two are the only derangements of the set. The number of possible derangements of a set is usually specified as \(!n\) (note, the exclamation is before the \(n\)). There exists an expansion of \(!n\) which is

$$

!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}

$$

Coming back to the probability question at hand, the number of ways by which all guests land up in different rooms from the ones they have…

Q: A large number of drunk guests arrive at a hotel where they have booked specific rooms. A careless receptionists hands over keys at random. What is the probability that at least one guest ends up in a room she booked?

Kindle, 6" E Ink Display, Wi-Fi - Includes Special Offers (Black)

A: A useful concept to understand is that of Derangement. A derangement is the number of ways a set can be permuted such that none of the elements are in their respective positions. For example if \(\{a,b,c\}\) is a set, then the derangements of the set are

\(\{b,c,a\}\)\(\{c,a,b\}\)The above two are the only derangements of the set. The number of possible derangements of a set is usually specified as \(!n\) (note, the exclamation is before the \(n\)). There exists an expansion of \(!n\) which is

$$

!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}

$$

Coming back to the probability question at hand, the number of ways by which all guests land up in different rooms from the ones they have…