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
[Math] envelopes probability of 3 letters
probability
probability
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
Best Answer
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 $n=8,r=3$.