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If $8$ letters are placed in $8$ adressed envelopes, then the probability that exactly $3$ letters goes to

correct envelopes and none of the remaining letters goes into correct envelopes is

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up vote 3 down vote accepted

The number of ways in which $n$ letters can be put in $n$ envelope so that no letter goes to its corresponding envelope is $$p_n=(n!)\sum_{k=0}^{n}\frac{(-1)^k}{k!}$$ Now $r$ letters which are to be placed in its corresponding envelope can be chosen in $n \choose r$ ways and the remaining $n-r$ letters can be put in $n-r$ envelopes so that no letter goes to its corresponding envelope is $p_{n-r}$. So our required probability is $$\frac{{n \choose r }(n-r)!\sum_{k=0}^{n-r}\frac{(-1)^k}{k!}}{n!}=\frac{1}{r!}\sum_{k=0}^{n-r}\frac{(-1)^k}{k!}$$ In your problem $n=8,r=3$.

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  • How many ways are there to choose which $3$ letters go into the correct envelopes?

  • The remaining $5$ letters must all go into the wrong envelopes. The number of ways to put these $5$ letters into the wrong envelopes is the number of derangements of a $5$-element set, sometimes written $!5$.

Combine these two numbers in the right way, and you have your answer.

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