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QUESTION:

N people arrive at a party all of whom are wearing hats. We collect all the hats and then redistribute them. What is the probability that no one gets their own hat?

ANSWER REASONING:

$$|\bigcup_{i=1}^N S_i|=\sum_{i=1}^N(-1)^{i+1}\binom{N}{i}(N-i)!=N!\sum_{i=1}^N\frac{(-1)^{i+1}}{i!}$$

What I don't get is the probability of exactly n men getting their own hats back.

So the reasoning given in the book is:P($A_{1} \bigcap A_{2} \bigcap A_{3}\bigcap......A_{n}$) = $N\choose n $ $(\frac{(N-n)!}{N!})$, where each $A_{i}$ is the probability of the i_th person choosing their own hat.

I am not sure why this makes sense because $(N-n)!$ gives us the number of ways in which (N-n) can choose their hats but does not exclude the possibility where (N-n)do not choose their own hats, which is what we want since we want only n people to be able to choose their hats, not the rest?

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    Assuming $A_1,A_2,\dots$ are the events that person $1$, person $2$, etc... gets their hats back, then you are misunderstanding what $\Pr(A_1\cap A_2\cap \dots \cap A_n)$ is. That is the probability that persons $1$ through $n$ get their hat back and possibly others among the remaining people also get their hats back. It is not the probability that exactly $n$ people get their hats back. Do not confuse $\Pr(A_1\cap A_2\cap \dots \cap A_n)$ with $\Pr(A_1\cap A_2\cap \dots \cap A_n\cap A_{n+1}^c \cap A_{n+2}^c\cap \dots \cap A_{N}^c)$ – JMoravitz Oct 01 '20 at 19:30
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    Also, the $\binom{N}{n}$ appears after adding together all $\binom{N}{n}$ different combinations of $n$ people. You should have had $\Pr(A_1\cap A_2\cap \dots \cap A_n) = \frac{1}{N}\cdot \frac{1}{N-1}\cdots \frac{1}{N-n+1}=\frac{(N-n)!}{N!}$. You just happen to have many terms in the sum that arose from inclusion-exclusion corresponding to the intersection of $n$ events who gave the same value who all get organized together to simplify the final sum. – JMoravitz Oct 01 '20 at 19:36
  • As for the probability you seem to actually be interested in (which I emphasize is not needed for this problem if you were to follow the book's inclusion-exclusion argument correctly) of the probability that exactly $n$ get their hat back and the remaining $N-n$ do not, that would be $\binom{N}{n}\times !(N-n)$ where the exclamationmark before the number refers to the number of derangements. – JMoravitz Oct 01 '20 at 19:39

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